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.
Practice A01 
\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}}\) 


\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}=\infty}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{(x^4 + 7x^2 + 3)} }\).
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 ).
Practice A01 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{(x^4+7x^2+3)}=\infty}\) 
Practice A02 
\(\displaystyle{\lim_{x\to\infty}{(x^53x^2+x21)}}\) 


\(\displaystyle{\lim_{x\to\infty}{(x^53x^2+x21)}=\infty}\) 
Evaluate \(\displaystyle{\lim_{x \to \infty}{(x^5  3x^2 + x  21)} }\).
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 \(\displaystyle{ \lim_{x \to \infty}{\left[ x^5( 1  3/x^3 + 1/x^4  21/x^5)\right]} = }\) \(\displaystyle{\infty(1  0 + 0  0) = (\infty)(1) = \infty }\)
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).
Practice A02 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{(x^53x^2+x21)}=\infty}\) 
Practice A03 
\(\displaystyle{\lim_{x\to\infty}{\frac{3x^2+5x+4}{x^3+7x}}}\) 

Practice A04 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x+5}{3x+7}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\) 
Evaluate \(\displaystyle{\lim_{x \to \infty}{\left[ \frac{x+5}{3x+7} \right]} }\).
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, \(x^1\).
\(\displaystyle{ \lim_{x \to \infty}{\left[ \left(\frac{x+5}{3x+7} \right) \cdot \frac{1/x}{1/x} \right]} = }\) \(\displaystyle{\lim_{x \to \infty}{\left[ \frac{1+5/x}{3+7/x} \right]} = \frac{1+0}{3+0} = 1/3 }\)
Practice A04 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\) 
Practice A05 
\(\displaystyle{\lim_{x\to\infty}{\frac{7}{x^316}}}\) 


\(\displaystyle{\lim_{x\to\infty}{\frac{7}{x^316}}=0}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{\frac{7}{x^316}} }\).
Using direct substitution, we get \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7}{x^316}\right]} = \frac{7}{\infty} = 0 }\)
Practice A05 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\frac{7}{x^316}}=0}\) 
Practice A06 
\(\displaystyle{\lim_{x\to\infty}{\frac{x^4+x}{5x^3+7}}}\) 

Practice A07 
\(\displaystyle{\lim_{x\to\infty}{\left[x\sqrt{x^2+9}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[x\sqrt{x^2+9}\right]}=\infty}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{\left[ x  \sqrt{x^2+9} \right]} }\).
Direct substitution yields \( \infty  \infty = \infty \) which is determinate.
Practice A07 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[x\sqrt{x^2+9}\right]}=\infty}\) 
Practice A08 
\(\displaystyle{\lim_{x\to\infty}{(3x^317x^2)}}\) 


\(\displaystyle{\lim_{x\to\infty}{(3x^317x^2)}=\infty}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{( 3x^317x^2 )} }\).
Direct substitution yields \(\displaystyle{ \lim_{x \to \infty}{( 3x^317x^2 )} = \infty  \infty }\), which is indeterminate. So we need to do some algebra.
\(\displaystyle{ \lim_{x \to \infty}{( 3x^317x^2 )} = }\) \(\displaystyle{\lim_{x \to \infty}{[(x^2)(3x17)]} = (\infty)(\infty) = \infty }\)
Practice A08 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{(3x^317x^2)}=\infty}\) 
Practice A09 
\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}}\) 


\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}=0}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{\frac{3}{x^2+5}} }\).
Using direct substitution, we get \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{3}{x^2+5}\right]} = \frac{3}{\infty} = 0 }\)
Practice A09 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\frac{3}{x^2+5}}=0}\) 
Practice A10 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^35x}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^35x}\right]}=7/2}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^3 + x + 12}{2x^3  5x} \right]} }\).
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.
Practice A10 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^35x}\right]}=7/2}\) 
Practice A11 
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2  3x + 12}{x^3 + 4x + 127} \right]} }\) 


\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2  3x + 12}{x^3 + 4x + 127} \right]}=0 }\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2  3x + 12}{x^3 + 4x + 127} \right]} }\).
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 \(x^3\).
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{(7x^2  3x + 12)}{(x^3 + 4x + 127)} \cdot \frac{1/x^3}{1/x^3} \right]} = }\) \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7/x  3/x^2 + 12/x^3}{1 + 4/x^2 + 127/x^3} \right]} = 0/1 = 0 }\)
In the last step, we used the Infinite Limits Theorem.
We could have used L'Hôpital's Rule, which would have been easier.
Practice A11 Final Answer 
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2  3x + 12}{x^3 + 4x + 127} \right]}=0 }\) 
Practice A12 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2+x+21}{11x}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2+x+21}{11x}\right]}=+\infty}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 + x + 21}{11x} \right]} }\).
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 \(x^2\).
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{(7x^2 + x + 21)}{(11x)} \cdot \frac{1/x^2}{1/x^2}\right]} = }\) \(\displaystyle{\lim_{x \to \infty}{\left[ \frac{7 + 1/x + 21/x^2}{11/x^21/x} \right]} = 7/0 = \infty }\)
In the last step, we used the Infinite Limits Theorem.
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.
Practice A12 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2+x+21}{11x}\right]}=+\infty}\) 
Practice A13 
\(\displaystyle{\lim_{x\to\infty}{\frac{4x^{10}+10000x^9}{5x^{10}+4}}}\) 

Practice A14 
\(\displaystyle{\lim_{x\to\infty}{\frac{3x^7}{5x^8+10x+2}}}\) 

Practice A15 
\(\displaystyle{\lim_{x\to\infty}{\frac{x^4}{x^3+5}}}\) 

Practice B01 
\(\displaystyle{\lim_{x\to\infty}{\frac{x+3}{\sqrt{x^2+4}}}}\) 

Practice B02 
\(\displaystyle{\lim_{x\to\infty}{\left[\sqrt{x^2+4x+1}x\right]}}\) 

Practice B03 
\(\displaystyle{ \lim_{x\to\infty}{\arctan(x)}}\) 


\(\displaystyle{ \lim_{x \rightarrow \infty}{\arctan(x)} = \pi/2 }\) 
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 stepbystep 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 \(\displaystyle{ \lim_{x \rightarrow \infty}{\arctan(x)} = \pi/2. } \)
Practice B03 Final Answer 
\(\displaystyle{ \lim_{x \rightarrow \infty}{\arctan(x)} = \pi/2 }\) 
Practice B04 
\(\displaystyle{\lim_{x\to\infty}{\left[x\sqrt{x^2+9}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[x\sqrt{x^2+9}\right]}=0}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{\left[ x  \sqrt{x^2+9} \right]} }\).
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).
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to \infty}{\left[ (x  \sqrt{x^2+9}) \frac{(x + \sqrt{x^2+9})}{(x + \sqrt{x^2+9})} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{(x  \sqrt{x^2+9})(x+\sqrt{x^2+9})}{x + \sqrt{x^2+9}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{x^2 + x\sqrt{x^2+9}  x\sqrt{x^2+9}  (x^2+9)}{x + \sqrt{x^2+9}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{x^2  (x^2+9)}{x + \sqrt{x^2+9}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{ 9}{x + \sqrt{x^2+9}} \right]}
\end{array} }\)
Now if we use direct substitution, we get \( 9/\infty = 0 \)
Practice B04 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[x\sqrt{x^2+9}\right]}=0}\) 
Practice B05 
\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}}\) 


\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}=1/2}\) 
Evaluate \(\displaystyle{ \lim_{x \to \infty}{\sqrt{ \frac{x^3 + 3x}{4x^3+7} }} }\).
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 \(x^{3/2}\).
\(\displaystyle{
\begin{array}{rcl}
\lim_{x \to \infty}{\left[ \sqrt{ \frac{x^3 + 3x}{4x^3+7} } \cdot \frac{x^{3/2}}{x^{3/2}}\right]} & = & \lim_{x \to \infty}{\sqrt{ \frac{(x^3 + 3x)}{(4x^3+7)} \cdot \frac{x^3}{x^3} }} \\\\
& = & \lim_{x \to \infty}{\sqrt{ \frac{(1 + 3/x^2)}{(4+7/x^3)} }} \\\\
& = & \sqrt{ \lim_{x \to \infty}{ \frac{(1 + 3/x^2)}{(4+7/x^3)} } } \\\\
& = & \sqrt{ \frac{1}{4} } = 1/2
\end{array}
}\)
In the last step, we used the Infinite Limits Theorem.
Practice B05 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\sqrt{\frac{x^3+3x}{4x^3+7}}}=1/2}\) 
Practice C01 
\(\displaystyle{\lim_{x\to\infty}{\frac{x^3}{\sqrt{x^6+4}}}}\) 
