L'Hôpital's rule helps us find many limits where direct substitution ends with the indeterminate forms 0/0 or ∞/∞. Review how (and when) it's applied.

What is L'Hôpital's rule?

L'Hôpital's rule helps us evaluate indeterminate limits of the form start fraction, 0, divided by, 0, end fraction or start fraction, infinity, divided by, infinity, end fraction.
In other words, it helps us find limxcu(x)v(x)\displaystyle\lim_{x\to c}\dfrac{u(x)}{v(x)}, where limxcu(x)=limxcv(x)=0\displaystyle\lim_{x\to c}u(x)=\lim_{x\to c}v(x)=0 (or, alternatively, where both limits are plus minus, infinity).
The rule essentially says that if the limit limxcu(x)v(x)\displaystyle\lim_{x\to c}\dfrac{u'(x)}{v'(x)} exists, then the two limits are equal:
limxcu(x)v(x)=limxcu(x)v(x)\displaystyle\lim_{x\to c}\dfrac{u(x)}{v(x)}=\displaystyle\lim_{x\to c}\dfrac{u'(x)}{v'(x)}
Want to learn more about L'Hôpital's rule? Check out this video.

Using L'Hôpital's rule to find limits of quotients

Let's find, for example, limx07xsin(x)x2+sin(3x)\displaystyle\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)}.
Substituting x, equals, 0 into start fraction, 7, x, minus, sine, left parenthesis, x, right parenthesis, divided by, x, start superscript, 2, end superscript, plus, sine, left parenthesis, 3, x, right parenthesis, end fraction results in the indeterminate form start fraction, 0, divided by, 0, end fraction. So let's use L’Hôpital’s rule.
=limx07xsin(x)x2+sin(3x)=limx0ddx[7xsin(x)]ddx[x2+sin(3x)]LHopitals rule=limx07cos(x)2x+3cos(3x)=7cos(0)2(0)+3cos(30)Substitution=2\begin{aligned} &\phantom{=}\displaystyle\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[7x-\sin(x)]}{\dfrac{d}{dx}[x^2+\sin(3x)]}\qquad\gray{\text{L'Hopital's rule}} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{7-\cos(x)}{2x+3\cos(3x)} \\\\ &=\dfrac{7-\cos(0)}{2(0)+3\cos(3\cdot0)}\qquad\gray{\text{Substitution}} \\\\ &=2 \end{aligned}
Note that we were only able to use L’Hôpital’s rule because the limit limx0ddx[7xsin(x)]ddx[x2+sin(3x)]\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[7x-\sin(x)]}{\dfrac{d}{dx}[x^2+\sin(3x)]} actually exists.
Problem 1.1
limx0ex12x=?\displaystyle\lim_{x\to 0}\dfrac{e^x-1}{2x}=?
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Using L'Hôpital's rule to find limits of exponents

Let's find, for example, limx0(1+2x)1sin(x)\displaystyle\lim_{x\to 0}(1+2x)^{^{\LARGE\frac{1}{\sin(x)}}}. Substituting x, equals, 0 into the expression results in the indeterminate form 1, start superscript, start superscript, infinity, end superscript, end superscript.
To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting y, equals, left parenthesis, 1, plus, 2, x, right parenthesis, start superscript, start superscript, start fraction, 1, divided by, sine, left parenthesis, x, right parenthesis, end fraction, end superscript, end superscript, we will find limx0ln(y)\displaystyle\lim_{x\to 0}\ln(y). Once we find it, we will be able to find limx0y\displaystyle\lim_{x\to 0}y.
natural log, left parenthesis, y, right parenthesis, equals, start fraction, natural log, left parenthesis, 1, plus, 2, x, right parenthesis, divided by, sine, left parenthesis, x, right parenthesis, end fraction
ln(y)=ln((1+2x)1sin(x))=1sin(x)ln(1+2x)Logarithm properties=ln(1+2x)sin(x)\begin{aligned} \ln(y)&=\ln\left((1+2x){^{\LARGE\frac{1}{\sin(x)}}}\right) \\\\ &=\dfrac{1}{\sin(x)}\ln(1+2x)\qquad\gray{\text{Logarithm properties}} \\\\ &=\dfrac{\ln(1+2x)}{\sin(x)} \end{aligned}
Substituting x, equals, 0 into start fraction, natural log, left parenthesis, 1, plus, 2, x, right parenthesis, divided by, sine, left parenthesis, x, right parenthesis, end fraction results in the indeterminate form start fraction, 0, divided by, 0, end fraction, so now it's L’Hôpital’s rule's turn to help us with our quest!
=limx0ln(y)=limx0ln(1+2x)sin(x)=limx0ddx[ln(1+2x)]ddx[sin(x)]LHopitals rule=limx0(21+2x)cos(x)=(21)1Substitution=2\begin{aligned} &\phantom{=}\displaystyle\lim_{x\to 0}\ln(y) \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\ln(1+2x)}{\sin(x)} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\ln(1+2x)]}{\dfrac{d}{dx}[\sin(x)]}\qquad\gray{\text{L'Hopital's rule}} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\left(\dfrac{2}{1+2x}\right)}{\cos(x)} \\\\ &=\dfrac{\left(\dfrac{2}{1}\right)}{1}\qquad\gray{\text{Substitution}} \\\\ &=2 \end{aligned}
We found that limx0ln(y)=2\displaystyle\lim_{x\to 0}\ln(y)=2, which means limx0y=e2\displaystyle\lim_{x\to 0}y=e^2.
Since natural log, left parenthesis, x, right parenthesis is continuous, we know that limx0ln(y)=ln(limx0y)\displaystyle\lim_{x\to 0}\ln(y)=\ln\left(\lim_{x\to 0}y\right), which means ln(limx0y)=2\ln\left(\displaystyle\lim_{x\to 0}y\right)=2:
ln(limx0y)=2eln(limx0y)=e2Raise e by both sideslimx0y=e2\begin{aligned} \ln\left(\displaystyle\lim_{x\to 0}y\right)&=2 \\\\ e^{\ln\left(\displaystyle\lim_{x\to 0}y\right)}&=e^2\qquad\gray{\text{Raise }e\text{ by both sides}} \\\\ \displaystyle\lim_{x\to 0}y&=e^2 \end{aligned}
Problem 2.1
limx0[cos(2πx)]1x=?\displaystyle\displaystyle\lim_{x\to0} [\cos\left(2\pi x \right)]^{^{\LARGE\frac 1 x}}=?
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Want to try more problems like this? Check out this exercise.