You CAN Ace Calculus
Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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This page discusses infinite Limits (or they may be called Limits At Infinity in your textbook) which refers to cases where the variable in question goes off to infinity. In limit notation, they look like \(\displaystyle{ \lim_{x \rightarrow \infty }{~f(x)} }\) or \(\displaystyle{ \lim_{x \rightarrow -\infty }{~f(x)} }\) . | ||||||||||||||||||
If your limit looks like \(\displaystyle{ \lim_{x \rightarrow c }{~f(x)} }\) where \(c\) as a finite number, then you need to go to the finite limits page. (The following panel explains the terminology and how 17calculus defines finite and infinite limits.) | ||||||||||||||||||
Finite Limits, Infinite Limits, Limits At Infinity . . . Terminology Explained
The use of the terms finite limits, infinite limits and limits at infinity are used differently in various books and your instructor may have their own idea of what they mean. In this panel, we will try to break down the cases and explain the various ways these terms can be used as well as how we use them here at 17calculus.
You can see where the confusion lies. The terms finite limits and infinite limits are used to mean two different things, referring to either \(c\) or \(L\). It is possible to have \(c = \infty\) and \(L\) be finite. So is this an infinite limit or a finite limit? It depends if you are talking about \(c\) or \(L\). |
When evaluating \(\displaystyle{ \lim_{x \to \pm \infty}{~f(x)} }\) you need to determine if the graph of the function is leveling off at a value ( and, if so, what that value is ) or if it is going off to infinity ( either \(+\infty\) or \(-\infty\) ). You don't want to try to figure it out off a graph. You need to do it mathematically (from the equation).
This is the main theorem you will use.
Infinite Limits Theorem |
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\(\displaystyle{ \lim_{x \rightarrow \pm \infty}{\left[ \frac{1}{x} \right]} = 0 }\) |
You can use the limit laws to apply this theorem to the case when you have \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{a}{x^k}\right]} }\) where \(k\) is a positive rational number and \(a\) is a real number. Here is an example. Try it on your own before looking at the solution.
Evaluate \(\displaystyle{ \lim_{x\to\infty}{\frac{3}{x^2}} }\).
\(\displaystyle{ \lim_{x\to\infty}{\frac{3}{x^2}} = 0 }\)
\(\displaystyle{ \lim_{x\to\infty}{\frac{3}{x^2}} = 3\lim_{x\to\infty}{\frac{1}{x^2}} }\) |
Constant Multiple Rule |
\(\displaystyle{ 3\left[ \lim_{x\to\infty}{\frac{1}{x}} \right] \left[ \lim_{x\to\infty}{\frac{1}{x}}\right] }\) |
Multiplication Rule |
\(\displaystyle{ 3(0)(0) = 0 }\) |
Infinite Limits Theorem |
Final Answer |
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\(\displaystyle{ \lim_{x\to\infty}{\frac{3}{x^2}} = 0 }\) |
close solution |
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new. |
Infinite Limits - Practice Problems Conversion |
[A01-1051] - [A02-1052] - [A03-1053] - [A04-1054] - [A05-1055] - [A06-1056] - [A07-1057] - [A08-1058] - [A09-1059] |
[A10-1060] - [A11-1061] - [A12-1062] - [A13-1063] - [A14-1064] - [A15-1065] |
[B01-1045] - [B02-1047] - [B03-1048] - [B04-1049] - [B05-1050] - [C01-1044] |
Please update your notes to this new numbering system. The display of this conversion information is temporary. |
Instructions - Unless otherwise instructed, evaluate the following limits, giving your answers in exact terms.
Basic Problems |
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\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}=\infty}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}}\)
Solution |
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Direct substitution yields
\(\displaystyle{ \lim_{x \to \infty}{(x^4 + 7x^2 + 3)} = \infty + \infty + 3 = \infty }\)
This form is determinate.
Note: Some textbooks and websites will give the answer as DNE (Does Not Exist). However, according to this definition, strictly this is not correct. Read your textbook carefully and ask your instructor what they expect (and make sure you understand why they expect a certain answer).
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}=\infty}\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}=\infty}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}}\)
Solution |
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Direct substitution yields \(\displaystyle{ \lim_{x \to \infty}{(x^5 - 3x^2 + x - 21)} = \infty - \infty + \infty - 21 }\)
This is indeterminate because of the \(\infty - \infty\). So we need to do some algebra to see if we can get a determinate form. There are several ways to do this. In this case, we will factor out the highest power of x to get \(\displaystyle{ \lim_{x \to \infty}{(x^5 - 3x^2 + x - 21)} = }\) \(\displaystyle{\lim_{x \to \infty}{\left[ x^5( 1 - 3/x^3 + 1/x^4 - 21/x^5)\right]} }\)
We know from the Infinite Limits Theorem that |
\(\displaystyle{ \lim_{x \to \infty}{1/x^3} = 0 }\) |
\(\displaystyle{ \lim_{x \to \infty}{1/x^4} = 0 }\) |
\(\displaystyle{ \lim_{x \to \infty}{1/x^5} = 0 }\) |
That leaves us with |
\(\displaystyle{ \lim_{x \to \infty}{\left[ x^5( 1 - 3/x^3 + 1/x^4 - 21/x^5)\right]} }\) |
\( \infty(1 - 0 + 0 - 0) \) |
\( (\infty)(1) = \infty \) |
Note: Some textbooks and websites will give the answer as DNE (Does Not Exist). However, according to the definition here, strictly this is not correct. Read your textbook carefully and ask your instructor what they expect (make sure you understand why they expect what they do).
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{(x^5-3x^2+x-21)}=\infty}\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{\frac{3x^2+5x+4}{x^3+7x}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\frac{3x^2+5x+4}{x^3+7x}}}\)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\)
Solution |
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Direct substitution yields \(\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{x+5}{3x+7} \right]} = \frac{-\infty}{-\infty}}\)
This is indeterminate. There are a couple of ways to work this. We will use algebra and the Infinite Limits Theorem here. (L'Hôpital's Rule would work here also.) To do this, we divide the numerator and denominator by the highest power of \(x\), in this case, \(x^1\).
\(\displaystyle{ \lim_{x \to -\infty}{\left[ \left(\frac{x+5}{3x+7} \right) \cdot \frac{1/x}{1/x} \right]} = }\) \(\displaystyle{\lim_{x \to -\infty}{\left[ \frac{1+5/x}{3+7/x} \right]} = \frac{1+0}{3+0} = 1/3 }\)
Final Answer |
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\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\) |
close solution |
\(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}=0}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}}\)
Solution |
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Using direct substitution, we get \(\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{7}{x^3-16}\right]} = \frac{7}{-\infty} = 0 }\)
Final Answer |
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\(\displaystyle{\lim_{x\to-\infty}{\frac{7}{x^3-16}}=0}\) |
close solution |
\(\displaystyle{\lim_{x\to-\infty}{\frac{x^4+x}{5x^3+7}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\frac{x^4+x}{5x^3+7}}}\)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}=-\infty}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}}\)
Solution |
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Direct substitution yields \( -\infty - \infty = -\infty \) which is determinate.
Final Answer |
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\(\displaystyle{\lim_{x\to-\infty}{\left[x-\sqrt{x^2+9}\right]}=-\infty}\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}=\infty}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}}\)
Solution |
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Direct substitution yields \(\displaystyle{ \lim_{x \to \infty}{( 3x^3-17x^2 )} = \infty - \infty }\), which is indeterminate. So we need to do some algebra.
\(\displaystyle{ \lim_{x \to \infty}{( 3x^3-17x^2 )} = }\) \(\displaystyle{\lim_{x \to \infty}{[(x^2)(3x-17)]} = (\infty)(\infty) = \infty }\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{(3x^3-17x^2)}=\infty}\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}=0}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}}\)
Solution |
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Using direct substitution, we get \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{3}{x^2+5}\right]} = \frac{3}{\infty} = 0 }\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}=0}\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}=7/2}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\)
Solution |
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Direct substitution yields \( \infty - \infty \) in the denominator, which is indeterminate. So we need to make some changes. Let's divide the numerator and denominator by the highest power of x, which in this case is \(x^3\).
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^3 + x + 12}{2x^3 - 5x} \cdot \frac{1/x^3}{1/x^3}\right]} = }\) \(\displaystyle{\lim_{x \to \infty}{\left[ \frac{7 + 1/x^2 + 12/x^3}{2 - 5/x^2} \right]} = }\) \(\displaystyle{\frac{7 + 0 + 0}{2 - 0} = 7/2 }\)
In the last step, we used the Infinite Limits Theorem.
If you know L'Hôpital's Rule, you could have used it here, which would have been easier.
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}=7/2}\) |
close solution |
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} }\)
Problem Statement |
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\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} }\)
Final Answer |
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\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]}=0 }\) |
Problem Statement |
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\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} }\)
Solution |
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Direct substitution yields \(\infty - \infty\) in the numerator, which is indeterminate. Let's divide the numerator and denominator by the highest power of x, in this case \(x^3\).
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{(7x^2 - 3x + 12)}{(x^3 + 4x + 127)} \cdot \frac{1/x^3}{1/x^3} \right]} }\) |
Use the Infinite Limits Theorem. |
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7/x - 3/x^2 + 12/x^3}{1 + 4/x^2 + 127/x^3} \right]} = 0/1 = 0 }\) |
We could have used L'Hôpital's Rule, which would have been easier.
Final Answer |
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\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]}=0 }\) |
close solution |
\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}=+\infty}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\)
Solution |
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Direct substitution yields \(\infty - \infty\) in the numerator which is indeterminate. So let's divide the numerator and denominator by the highest power, which is \(x^2\).
\(\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{(7x^2 + x + 21)}{(11-x)} \cdot \frac{1/x^2}{1/x^2}\right]} }\) |
Use the Infinite Limits Theorem |
\(\displaystyle{\lim_{x \to -\infty}{\left[ \frac{7 + 1/x + 21/x^2}{11/x^2-1/x} \right]} = 7/0 = \infty }\) |
Now, the tendency here is to stop and say the answer is \(+\infty\). But since x goes to \(-\infty\), we need to stop and take the negative sign into account. In this case, since we divided the numerator and denominator by \(x^2\) which is always positive, then the sign doesn't change. However, if the highest power would have been odd, the sign would have changed on both the numerator and denominator leaving the limit \(-\infty\).
We could have used L'Hôpital's Rule, which would have been easier.
Final Answer |
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\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}=+\infty}\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{\frac{4x^{10}+10000x^9}{5x^{10}+4}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\frac{4x^{10}+10000x^9}{5x^{10}+4}}}\)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{\lim_{x\to\infty}{\frac{3x^7}{5x^8+10x+2}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\frac{3x^7}{5x^8+10x+2}}}\)
Solution |
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close solution |
\(\displaystyle{\lim_{x\to\infty}{\frac{x^4}{x^3+5}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\frac{x^4}{x^3+5}}}\)
Solution |
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video by PatrickJMT
close solution |
Intermediate Problems |
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\(\displaystyle{\lim_{x\to\infty}{\frac{x+3}{\sqrt{x^2+4}}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\frac{x+3}{\sqrt{x^2+4}}}}\)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{ \lim_{x \to \pm \infty}{\frac{3x^4+5x}{2x^4-3x^3+10}} }\)
Problem Statement |
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\(\displaystyle{ \lim_{x \to \pm \infty}{\frac{3x^4+5x}{2x^4-3x^3+10}} }\)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{ \lim_{x\to\infty}{\arctan(x)}}\)
Problem Statement |
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\(\displaystyle{ \lim_{x\to\infty}{\arctan(x)}}\)
Final Answer |
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\(\displaystyle{ \lim_{x \to \infty}{\arctan(x)} = \pi/2 }\) |
Problem Statement |
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\(\displaystyle{ \lim_{x\to\infty}{\arctan(x)}}\)
Solution |
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The question was to determine the limit \(\displaystyle{ \lim_{x \rightarrow \infty}{\arctan(x)} }\).
Now, without studying this, I wouldn't know off the top of my head what this limit is or even if it existed. In fact, the technique to determine this is a little unusual. So, I am going to take you step-by-step through a procedure that you need to start doing as you work through problems.
The first question I asked myself was, what technique that I already know, can I use to determine this? Well, I could graph it and get an idea of an answer. If you graph this, it looks like the graph is leveling off to the right and that a limit might be about 1.5. However, we can't tell if it is just rising very slowly or if a limit does actually exist. We need to find an exact answer.
Whenever I have an inverse trig function, one technique I try is to convert to regular trig functions since I am much more familiar with them. So we have
\(\displaystyle{
\begin{align}
\theta &= \arctan(x) \\
tan( \theta ) &= x \\
\frac{\sin( \theta)}{\cos( \theta)} &= x
\end{align}
}\)
It's a good idea to use \( \theta \) (or another Greek letter) to remind yourself that inverse trig functions return angles. Also remember for inverse trig functions to make sense, we must restrict the domain of the corresponding trig function. For this case,
we restrict the domain of the tangent function to \( -\pi/2 < \theta < \pi/2 \). [ This is something you need to keep in mind whenever you work with inverse trig functions. ]
So now we have enough information to determine the limit. The last equation is \( \displaystyle{ x=\frac{\sin( \theta )}{\cos( \theta )} }\) and we know that \( x \rightarrow \infty \). So we need to know the angle that would cause x to go off to infinity. Do you see it now?
I would be \( \theta=\pi/2 \) since the numerator is 1 and the denominator is zero. (Why not \( \theta=-\pi/2 \)?)
Therefore our limit is \(\displaystyle{ \lim_{x \rightarrow \infty}{\arctan(x)} = \pi/2. } \)
Final Answer |
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\(\displaystyle{ \lim_{x \to \infty}{\arctan(x)} = \pi/2 }\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}=0}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}}\)
Solution |
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Direct substitution yields \(\infty - \infty\) which is indeterminate. The technique we will use here is called rationalizing. We multiply the numerator and denominator by the conjugate of \(x - \sqrt{x^2+9} \). The conjugate is the same expression with one of the signs changed (it doesn't matter which sign).
\(\displaystyle{ \lim_{x \to \infty}{\left[ (x - \sqrt{x^2+9}) \frac{(x + \sqrt{x^2+9})}{(x + \sqrt{x^2+9})} \right]} }\) |
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{(x - \sqrt{x^2+9})(x+\sqrt{x^2+9})}{x + \sqrt{x^2+9}} \right]} }\) |
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{x^2 + x\sqrt{x^2+9} - x\sqrt{x^2+9} - (x^2+9)}{x + \sqrt{x^2+9}} \right]} }\) |
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{x^2 - (x^2+9)}{x + \sqrt{x^2+9}} \right]} }\) |
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{ -9}{x + \sqrt{x^2+9}} \right]} }\) |
Now if we use direct substitution, we get \( -9/\infty = 0 \)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{\left[x-\sqrt{x^2+9}\right]}=0}\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}}\)
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}=1/2}\) |
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}}\)
Solution |
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Direct substitution yields \(\infty / \infty \) which is indeterminate. So we need to use some algebra. Let's divide the numerator and denominator by the highest power of x, in this case \(x^{3/2}\).
\(\displaystyle{ \lim_{x \to \infty}{\left[ \sqrt{ \frac{x^3 + 3x}{4x^3+7} } \cdot \frac{x^{3/2}}{x^{3/2}}\right]} }\) |
\(\displaystyle{ \lim_{x \to \infty}{\sqrt{ \frac{(x^3 + 3x)}{(4x^3+7)} \cdot \frac{x^3}{x^3} }} }\) |
\(\displaystyle{ \lim_{x \to \infty}{\sqrt{ \frac{(1 + 3/x^2)}{(4+7/x^3)} }} }\) |
\(\displaystyle{ \sqrt{ \lim_{x \to \infty}{ \frac{(1 + 3/x^2)}{(4+7/x^3)} } } }\) |
Use the Infinite Limits Theorem |
\(\displaystyle{ \sqrt{ \frac{1}{4} } = 1/2 }\) |
Final Answer |
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\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}=1/2}\) |
close solution |
\(\displaystyle{\lim_{x\to\infty}{\left[\sqrt{x^2+4x+1}-x\right]}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to\infty}{\left[\sqrt{x^2+4x+1}-x\right]}}\)
Solution |
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video by Khan Academy
close solution |
Advanced Problems |
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\(\displaystyle{\lim_{x\to-\infty}{\frac{x^3}{\sqrt{x^6+4}}}}\)
Problem Statement |
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\(\displaystyle{\lim_{x\to-\infty}{\frac{x^3}{\sqrt{x^6+4}}}}\)
Solution |
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video by PatrickJMT
close solution |