Practice A01 |
\(\displaystyle{\lim_{x\to\pi/2}{\left[\frac{\tan(2x)}{x-\pi/2}\right]}}\) |
If we plug in \( \pi/2 \) for x, we get \( 0/0 \) which is indeterminate. This means we can use L'Hôpital's Rule.
\(\displaystyle{ \lim_{x \to \pi/2}{\left[ \frac{\tan(2x)}{x-\pi/2} \right]} = }\)
\(\displaystyle{\lim_{x \to \pi/2}{\left[ \frac{2 \sec^2(2x)}{1} \right]} = 2 \sec^2(\pi) = 2 }\)
Note: You could also solve this using trig identities (but that's a lot more work).
Practice A01 Final Answer |
\(\displaystyle{\lim_{x\to\pi/2}{\left[\frac{\tan(2x)}{x-\pi/2}\right]}=2}\) |
Practice A02 |
\(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}\) |
First of all, you always want to try direct substitution. When you do this, you get \(0/0\) which is indeterminate.
There are at least two ways to solve this.
We will solve it using L'Hôpital's Rule here. See the
finite limits page for an alternative solution.
\(\displaystyle{ \lim_{x \to 2} \left[ \frac{3x^2-x-10}{x^2-4} \right] = }\) \(\displaystyle{
\lim_{x \to 2} \left[ \frac{6x-1}{2x} \right] = }\) \(\displaystyle{
\frac{6(2)-1}{2(2)} = 11/4 }\)
Remember: You can use L'Hôpital's Rule only if you have an indeterminate form.
Practice A02 Final Answer |
\(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}=11/4}\) |
Practice A03 |
\(\displaystyle{\lim_{x\to3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}\) |
First of all, you always want to try direct substitution. When you do this, you get \(0/0\) which is indeterminate.
There are at least two ways to solve this.
We will solve it using L'Hôpital's Rule here. See the finite limits page for an alternative solution.
\(\displaystyle{
\lim_{x \to 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] } = }\) \(\displaystyle{
\lim_{x \to 3}{ \left[\frac{4x^3}{4x-5}\right] } = \frac{4(3)^3}{4(3)-5} = 108/7 }\)
Remember: You can use L'Hôpital's Rule only if you have an indeterminate form.
Practice A03 Final Answer |
\(\displaystyle{\lim_{x\to3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}=108/7}\) |
Practice A04 |
\(\displaystyle{\lim_{x\to27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}\) |
Of course, the first thing we want to try is to plug in 27 directly for x. This gives us \(0/0\) which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to 27}{\left[ \frac{x - 27}{x^{1/3} - 3}\right]} \\\\
& = & \lim_{x \to 27}{ \frac{1}{(1/3)x^{-2/3}} } \\\\
& = & \lim_{x \to 27}{ \frac{x^{2/3}}{1/3} } \\\\
& = & \lim_{x \to 27}{ 3x^{2/3} } \\\\
& = & 3(27)^{2/3} = 3(9) = 27
\end{array}
}\)
Note - - We could have used algebra and factoring to solve this problem. See that solution on the finite limits page.
Practice A04 Final Answer |
\(\displaystyle{\lim_{x\to27}{\left[\frac{x-27}{x^{1/3}-3}\right]}=27}\) |
Practice A05 |
\(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}\) |
Direct substitution yields \(0/0\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to 1}{\left[ \frac{x^{1/3}-1}{x^{1/4}-1}\right] } \\\\
& = & \lim_{x \to 1}{ \frac{(1/3)x^{-2/3}}{(1/4)x^{-3/4}} } \\\\
& = & \lim_{x \to 1}{ \frac{4}{3} \frac{x^{3/4}}{x^{2/3}} } \\\\
& = & \lim_{x \to 1}{ \frac{4}{3} x^{3/4 - 2/3} } \\\\
& = & \lim_{x \to 1}{ \frac{4}{3} x^{9/12 - 8/12} } \\\\
& = & \lim_{x \to 1}{ \frac{4}{3} x^{1/12} } = \frac{4}{3} 1^{1/12} = \frac{4}{3}
\end{array}
}\)
Note - - We could have used algebra and factoring to solve this problem. See that solution on the finite limits page.
Practice A05 Final Answer |
\(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}=4/3}\) |
Practice A06 |
\(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}}\) |
Direct substitution yields \(0/0\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to 0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]} \\\\
& = & \lim_{x \to 0}{\left[\frac{1+\sec^2(x)}{\cos(x)}\right]} \\\\
& = & \lim_{x \to 0}{\left[\frac{1+[1/\cos^2(x)]}{\cos(x)}\right]} \\\\
& = & \lim_{x \to 0}{\left[\frac{\cos^2(x)+1}{\cos^3(x)}\right]} \\\\
& = & \frac{1+1}{1} = 2
\end{array}
}\)
After applying L'Hôpital's Rule once, the rest is just trig and algebra.
Practice A06 Final Answer |
\(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}=2}\) |
Practice A07 |
\(\displaystyle{\lim_{z\to\pi}{\left[\frac{z-\pi}{\sin(z)}\right]}}\) |
Direct substitution yields \(0/0\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{z \to \pi}{\left[ \frac{z - \pi}{sin(z)} \right]} = }\) \(\displaystyle{
\lim_{z \to \pi}{\left[ \frac{1}{cos(z)} \right]} = }\) \(\displaystyle{
\frac{1}{-1} = -1 }\)
Practice A07 Final Answer |
\(\displaystyle{\lim_{z\to\pi}{\left[\frac{z-\pi}{sin(z)}\right]}=-1}\) |
Practice A08 |
\(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}}\) |
Direct substitution yields \(0/0\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{y \to 0}{\left[ \frac{\sin(3y)}{\sin(4y)} \right]} = }\) \(\displaystyle{
\lim_{y \to 0}{\left[ \frac{3\cos(3y)}{4\cos(4y)} \right]} = \frac{3(1)}{4(1)} = 3/4 }\)
Note - - There is another way to work this using the identity \(\displaystyle{ \lim_{x \to 0}{\left[ \frac{\sin(x)}{x} \right]} = 1 }\).
Practice A08 Final Answer |
\(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}=3/4}\) |
Practice A09 |
\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\) |
Direct substitution yields
\(\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{x+5}{3x+7} \right]} = \frac{-\infty}{-\infty}}\)
This is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{x \to -\infty}{\left[ \frac{x+5}{3x+7} \right]} = \lim_{x \to -\infty}{ \frac{1}{3} } = 1/3}\)
Note - - This is the easiest way to work this problem. If you don't know L'Hôpital's Rule, you can use algebra. You will find that solution on the infinite limits page.
Practice A09 Final Answer |
\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\) |
Practice A10 |
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\) |
Direct substitution yields \( \infty - \infty \) in the denominator, which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{x \to \infty}{\left[ \frac{7x^3 + x + 12}{2x^3 - 5x} \right]} = \lim_{x \to \infty}{ \frac{21x^2 + 1}{6x^2} }
}\)
Direct substitution still gives us an indeterminate form, \(\infty / \infty\). So we can use L'Hôpital's Rule again.
\(\displaystyle{\lim_{x \to \infty}{ \frac{21x^2 + 1}{6x^2} } = }\) \(\displaystyle{
\lim_{x \to \infty}{ \frac{42x}{12x} } = \frac{42}{12} = 7/2}\)
Note - - This is the easiest way to work this problem. If you don't know L'Hôpital's Rule, you can use algebra. You will find that solution on the infinite limits page.
Practice A10 Final Answer |
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}=7/2}\) |
Practice A11 |
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2-3x+12}{x^3+4x+127}\right]}}\) |
Direct substitution yields \(\displaystyle{ \frac{\infty - \infty + 12}{\infty + \infty + 127} }\) which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} = }\) \(\displaystyle{
\lim_{x \to \infty}{\left[ \frac{14x - 3}{3x^2 + 4} \right]}}\)
Again, after direct substitution, we get \( \infty / \infty \), which is also indeterminate. Use L'Hôpital's Rule again.
\(\displaystyle{\lim_{x \to \infty}{\left[ \frac{14x - 3}{3x^2 + 4} \right]} = }\) \(\displaystyle{
\lim_{x \to \infty}{ \frac{14}{6x} } = \frac{14}{\infty} = 0}\)
Note - - This is the easiest way to work this problem. If you don't know L'Hôpital's Rule, you can use algebra. You will find that solution on the infinite limits page.
Practice A11 Final Answer |
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2-3x+12}{x^3+4x+127}\right]}=0}\) |
Practice A12 |
\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\) |
Direct substitution yields \(\infty / \infty\) which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{x \to -\infty}{\left[ \frac{7x^2 + x + 21}{11-x} \right]} = }\) \(\displaystyle{
\lim_{x \to -\infty}{ \frac{14x + 1}{-1} }}\)
Now, let's try direct substitution again.
\(\displaystyle{\lim_{x \to -\infty}{ \frac{14x + 1}{-1} } = \frac{-\infty + 1}{-1} = \infty}\)
Note - - This is the easiest way to work this problem. If you don't know L'Hôpital's Rule, you can use algebra. You will find that solution on the infinite limits page.
Practice A12 Final Answer |
\(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}=+\infty}\) |
Practice A13 |
\(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{x}}}\) |
Practice A14 |
\(\displaystyle{\lim_{x\to\infty}{\frac{\ln(x)}{x}}}\) |
Practice A15 |
\(\displaystyle{\lim_{x\to0^+}{\frac{\ln(x)}{x}}}\) |
Practice A16 |
\(\displaystyle{\lim_{x\to\infty}{\frac{x}{\ln(1+2e^x)}}}\) |
Practice A17 |
\(\displaystyle{\lim_{x\to\infty}{(x^2-x)}}\) |
Practice B01 |
\(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)-x}{\sin(x)-x}\right]}}\) |
Direct substitution yields \(0/0\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to 0}{\left[ \frac{\tan(x)-x}{\sin(x)-x} \right]} \\\\
& = & \lim_{x \to 0}{\left[ \frac{\sec^2(x)-1}{\cos(x)-1} \right]} \\\\
& = & \lim_{x \to 0}{\left[ \frac{2\sec(x)[\sec(x)\tan(x)]}{-\sin(x)} \right]} \\\\
& = & \lim_{x \to 0}{\left[ \frac{2\sec^2(x)\tan(x)}{-\sin(x)} \right]} \\\\
& = & \lim_{x \to 0}{\left[ \frac{-2\sin(x) }{\cos^2(x)\sin(x)\cos(x) } \right]} \\\\
& = & \lim_{x \to 0}{\left[ \frac{-2}{\cos^3(x)} \right]} \\\\
& = & \frac{-2}{1} = -2
\end{array}}\)
Practice B01 Final Answer |
\(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)-x}{\sin(x)-x}\right]}=-2}\) |
Practice B02 |
\(\displaystyle{\lim_{n\to\infty}{n^{2/n}}}\) |
Direct substitution yields \( \infty^0 \) which is indeterminate. We can't directly use L'Hôpital's Rule because L'Hôpital's Rule requires a fraction. So we have to use a trick.
Let's call the limit, \(y\), i.e. \(\displaystyle{ y = \lim_{n \to \infty}{n^{2/n}} }\) and take the natural log of both sides. This yields
\(\displaystyle{
\begin{array}{rcl}
\ln(y) & = & \lim_{n \to \infty}{\ln(n^{2/n})} \\\\
& = & \lim_{n \to \infty}{\frac{2\ln(n)}{n}} \\\\
& = & \lim_{n \to \infty}{\frac{2/n}{1}} = 0 \\\\
\ln(y) & = & 0 \\\\
e^{\ln(y)} & = & e^0 \\\\
y & = & 1
\end{array}
}\)
Practice B02 Final Answer |
\(\displaystyle{\lim_{n\to\infty}{n^{2/n}}=1}\) |
Practice B03 |
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x^5}{e^{5x}}\right]}}\) |
If we try direct substitution first, we get \( \infty/\infty \) which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to \infty}{\left[ \frac{x^5}{e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{5x^4}{5e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{x^4}{e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{4x^3}{5e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{12x^2}{25e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{24x}{125e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{24}{625e^{5x}} \right]} \\\\
& = & \frac{24}{\infty} = 0
\end{array}}\)
Note - - In this solution we used L'Hôpital's Rule five times until we no longer had an indeterminate form. Not shown here is the fact that, before each use of L'Hôpital's Rule, we tested for indeterminate forms by direct substitution.
Practice B03 Final Answer |
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x^5}{e^{5x}}\right]}=0}\) |
Practice B04 |
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{k}}}\) where k is a positive constant |
\(\displaystyle{
\begin{array}{rcl}
y & = & \lim_{n \to \infty}{ \sqrt[n]{k} } = \lim_{n \to \infty}{ k^{1/n} } \\\\
\ln(y) & = & \ln\left[ \lim_{n \to \infty}{ k^{1/n} } \right] \\\\
& = & \lim_{n \to \infty}{ \ln\left( k^{1/n} \right) } \\\\
& = & \lim_{n \to \infty}{ (1/n)\ln(k) } \\\\
& = & \lim_{n \to \infty}{ \frac{\ln(k)}{n} } \\\\
& = & 0 \\\\
e^{\ln(y)} & = & e^0 \\\\
y & = & 1
\end{array}}\)
Practice B04 Final Answer |
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{k}}=1}\) |
Practice B05 |
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{n^k}}}\) where k is a positive constant |
\(\displaystyle{
\begin{array}{rcl}
y & = & \lim_{n \to \infty}{ \sqrt[n]{n^k} } = \lim_{n \to \infty}{ n^{k/n} } \\\\
\ln(y) & = & \ln \left[ \lim_{n \to \infty}{ n^{k/n} } \right] \\\\
& = & \lim_{n \to \infty}{ \ln \left[ n^{k/n} \right] } \\\\
& = & \lim_{n \to \infty}{ (k/n)\ln(n) } \\\\
& = & \lim_{n \to \infty}{ \frac{k\ln(n)}{n} } \\\\
& = & \lim_{n \to \infty}{ \frac{k(1/n)}{1} } \\\\
& = & \lim_{n \to \infty}{ \frac{k}{n} } \\\\
& = & 0 \\\\
e^{\ln(y)} & = & e^0 \\\\
y & = & 1
\end{array}}\)
Practice B05 Final Answer |
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{n^k}}=1}\) |
Practice B06 |
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{\ln(n)}}}\) |
\(\displaystyle{
\begin{array}{rcl}
y & = & \lim_{n \to \infty}{ \sqrt[n]{\ln(n)} } = \lim_{n \to \infty}{ (\ln(n))^{1/n} } \\\\
\ln(y) & = & \ln \left[ \lim_{n \to \infty}{(\ln(n))^{1/n} } \right] \\\\
& = & \lim_{n \to \infty}{ \ln \left[ (\ln(n))^{1/n} \right] } \\\\
& = & \lim_{n \to \infty}{ (1/n)\ln(\ln(n)) } \\\\
& = & \lim_{n \to \infty}{ \frac{\ln(\ln(n))}{n} } \\\\
& = & \lim_{n \to \infty}{ \frac{1/n}{\ln(n)(1)} } \\\\
& = & \lim_{n \to \infty}{ \frac{1}{n\ln(n)} } \\\\
& = & 0 \\\\
e^{\ln(y)} & = & e^0 \\\\
y & = & 1
\end{array}}\)
Practice B06 Final Answer |
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{\ln(n)}}=1}\) |
Practice B07 |
\(\displaystyle{\lim_{x\to0^+}{\sqrt{x}\ln(x)}}\) |
Practice B08 |
\(\displaystyle{\lim_{x\to0}{\cot(2x)\sin(6x)}}\) |
Practice B09 |
\(\displaystyle{\lim_{x\to\infty}{(xe^{1/x}-x)}}\) |
Practice B10 |
\(\displaystyle{\lim_{x\to1}{\left[\frac{1}{\ln(x)}-\frac{1}{x-1}\right]}}\) |
Practice B11 |
\(\displaystyle{\lim_{x\to\infty}{(e^x+x)^{1/x}}}\) |
Practice B12 |
\(\displaystyle{\lim_{x\to0}{(\cos(3x))^{5/x}}}\) |
Practice C01 |
\(\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } }\) |
Hint: Use the substitution \(u=1/x\) and move the exponential to the denominator.
Direct substitution yields \(0/0\) which is indeterminate. So our next logical step is to use L'Hôpital's Rule. However, when we do that, we get an even more complicated result.
If we use the hint given above, the limit will approach \(-\infty\) and we have
\(\displaystyle{ \lim_{u \to -\infty}{ \frac{-e^{u}}{1/u} } = \lim_{u \to -\infty}{ \frac{-u}{e^{-u}} } }\).
Now we can use L'Hôpital's Rule (since direct substitution yields the indeterminate fraction \(\infty/\infty\)) to get
\(\displaystyle{ \lim_{u \to -\infty}{ \frac{-u}{e^{-u}} } = }\)
\(\displaystyle{ \lim_{u \to -\infty}{ \frac{-1}{-e^{-u}} } = }\)
\(\displaystyle{ \frac{1}{e^{\infty}} = 0 }\)
Practice C01 Final Answer |
\(\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } = 0 }\) |