L’Hopital’s Rule

L’Hopital’s Rule
More Practice

L’Hôpital’s Rule is one of those quirky things to remember in Calculus when you are taking limits (like we did in the Limits and Continuity section). The Calculus books don’t seem to introduce this rule until the Integration Techniques sections, so here it is!

When taking limits, you may end up with what we call indeterminate forms, which are $ \displaystyle \frac{0}{0}$ or $ \displaystyle \frac{\infty }{\infty }$. The reason these fractions are called indeterminate is because we don’t know if a limit exists, or if one exists, we don’t know what it is. Earlier we learned to try some algebra on these (factoring and crossing out, or multiplying by conjugates, for example) to try to find the limit.

Basically what L’Hôpital’s Rule says is, in certain cases, to get the limit of a quotient, you can just take the limit of a quotient of the derivatives of the numerator (top) and denominator (bottom). Here is a more formal definition:

L’Hôpital’s Rule:

If $ f$ and $ g$ are differentiable functions and if $ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\frac{{f\left( x \right)}}{{g\left( x \right)}}$ is $ \displaystyle \frac{0}{0}$ or $ \displaystyle \frac{\infty }{\infty }$ or $ \displaystyle \frac{{-\infty }}{{-\infty }}$ or $ \displaystyle \frac{{-\infty }}{\infty }$ or $ \displaystyle \frac{\infty }{{-\infty }}$, then

$ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\frac{{f\left( x \right)}}{{g\left( x \right)}}=\underset{{x\to c}}{\mathop{{\lim }}}\frac{{{f}’\left( x \right)}}{{{g}’\left( x \right)}}$ (as long as the limit on the right exists or is infinite).

What this says is that if you have two functions whose limits fit the conditions above, you can take the derivative of the top function and bottom function over and over again until you get limits that “work”.

Remember, with L’Hôpital’s, when you work with quotients, you don’t need the quotient rule; you just take the derivative of the top (numerator) and bottom (denominator), and then take the limit. If you can’t get the limit after taking the derivatives of the top and bottom, take the derivative again and then try (you can do this indefinitely!) To check these, you can always put them in the calculator, putting in a very small number (if limit goes to 0) or very large number (if limit goes to $ \infty $) to see if you get a close answer.

Here are some examples, using both algebraic techniques and L’Hôpital’s.

Find Limits	Algebraic Solution	L’Hôpital’s Rule Solution
$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3x+2}}{{5x-1}}$ Indeterminate Form: $ \displaystyle \frac{\infty }{\infty }$	Divide each term by $ x$, since this is the highest degree (exponent). Remember that $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{1}{x}=0$: $ \displaystyle \begin{align}\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3x+2}}{{5x-1}}&=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{{3x}}{x}+\frac{2}{x}}}{{\frac{{5x}}{x}-\frac{1}{x}}}\,\,\\&=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\,\,\frac{{3+\frac{2}{x}}}{{5-\frac{1}{x}}}=\frac{3}{5}\end{align}$	Take the derivative of the top and bottom of the fraction, and then take the limit: $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3x+2}}{{5x-1}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{3}{5}\,\,=\frac{3}{5}$ Since there is no $ x$ in the answer, ignore $ x$ going to $ \infty $. Much easier!
$ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{{3x}}$ Indeterminate Form: $ \displaystyle \frac{0}{0}$	Get this in the form $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sin x}}{x}$ (move a $ \displaystyle \frac{1}{3}$ to the outside, and then multiply by $ \displaystyle \frac{4}{4}$) since $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sin x}}{x}=1$: $ \displaystyle \begin{align}\underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{{3x}}=\frac{1}{3}\cdot \underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{x}=\frac{1}{3}\cdot \frac{4}{4}\cdot \underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{x}\\=\,\frac{1}{3}\cdot \frac{4}{1}\underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{{4x}}=\frac{4}{3}\cdot 1=\frac{4}{3}\end{align}$	Take the derivative of the top and bottom of the fraction, and then take the limit: $ \displaystyle \begin{align}\underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{{3x}}&=\underset{{x\to 0}}{\mathop{{\lim }}}\frac{{4\cos 4x}}{3}\\&=\frac{{4\cos \left( 0 \right)}}{3}=\frac{{4\cdot 1}}{3}=\frac{4}{3}\end{align}$
$ \displaystyle \underset{{x\to -\infty }}{\mathop{{\lim }}}\,\frac{x}{{{{x}^{2}}-1}}$ Indeterminate Form: $ \displaystyle \frac{0}{0}$	Divide each term by $ {{x}^{2}}$, since this is the highest degree (exponent). Remember that $ \displaystyle \underset{{x\to -\infty }}{\mathop{{\lim }}}\,\frac{1}{x}=0$: $ \displaystyle \begin{align}\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\,\frac{x}{{{{x}^{2}}-1}}&=\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\,\frac{{\frac{x}{{{{x}^{2}}}}}}{{\frac{{{{x}^{2}}}}{{{{x}^{2}}}}-\frac{1}{{{{x}^{2}}}}}}\\&=\underset{{x\to \,-\infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{x}}}{{1-\frac{1}{{{{x}^{2}}}}}}=\frac{0}{{1-0}}=0\end{align}$	Take the derivative of the top and bottom of the fraction (do this twice!), and then take the limit: $ \displaystyle \begin{align}\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\frac{x}{{{{x}^{2}}-1}}&=\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\frac{1}{{2x}}\\&=\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\frac{0}{2}=0\end{align}$ Since there is no $ x$ in the answer, ignore $ x$ going to $ -\infty $.
$ \displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sqrt[3]{{h+27}}-3}}{h}$ Indeterminate Form: $ \displaystyle \frac{0}{0}$	Too complicated to do algebraically!	Remember that you don’t need the quotient rule; just take derivative on top and bottom: $ \begin{align}\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sqrt[3]{{h+27}}-3}}{h}\,&=\,\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{{{{\left( {h+27} \right)}}^{{\frac{1}{3}}}}-3}}{h}=\,\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\frac{1}{3}{{{\left( {h+27} \right)}}^{{-\frac{2}{3}}}}-0}}{1}\\&=\,\frac{1}{3}{{\left( {0+27} \right)}^{{-\frac{2}{3}}}}=\frac{1}{3}{{\left( {\frac{1}{{27}}} \right)}^{{\frac{2}{3}}}}=\frac{1}{3}\left( {\frac{1}{9}} \right)=\frac{1}{{27}}\end{align}$
$ \displaystyle \underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {\pi +t} \right)}}{t}$ Indeterminate Form: $ \displaystyle \frac{0}{0}$	Not sure how to do this algebraically!	Take the derivative of the top and bottom of the fraction, and then take the limit: $ \displaystyle \begin{align}\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\sin \left( {\pi +t} \right)}}{t}&=\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\cos \left( {\pi +t} \right)}}{1}\\&=\cos \left( {\pi +0} \right)=-1\end{align}$

Find Limits

Algebraic Solution

L’Hôpital’s Rule Solution

$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3x+2}}{{5x-1}}$

Indeterminate Form: $ \displaystyle \frac{\infty }{\infty }$

Divide each term by $ x$, since this is the highest degree (exponent). Remember that $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{1}{x}=0$:

$ \displaystyle \begin{align}\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3x+2}}{{5x-1}}&=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{{3x}}{x}+\frac{2}{x}}}{{\frac{{5x}}{x}-\frac{1}{x}}}\,\,\\&=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\,\,\frac{{3+\frac{2}{x}}}{{5-\frac{1}{x}}}=\frac{3}{5}\end{align}$

Take the derivative of the top and bottom of the fraction, and then take the limit:

$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{3x+2}}{{5x-1}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{3}{5}\,\,=\frac{3}{5}$

Since there is no $ x$ in the answer, ignore $ x$ going to $ \infty $. Much easier!

$ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{{3x}}$

Indeterminate Form: $ \displaystyle \frac{0}{0}$

Get this in the form $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sin x}}{x}$ (move a $ \displaystyle \frac{1}{3}$ to the outside, and then multiply by $ \displaystyle \frac{4}{4}$) since $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sin x}}{x}=1$:

$ \displaystyle \begin{align}\underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{{3x}}=\frac{1}{3}\cdot \underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{x}=\frac{1}{3}\cdot \frac{4}{4}\cdot \underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{x}\\=\,\frac{1}{3}\cdot \frac{4}{1}\underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{{4x}}=\frac{4}{3}\cdot 1=\frac{4}{3}\end{align}$

Take the derivative of the top and bottom of the fraction, and then take the limit:

$ \displaystyle \begin{align}\underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{{3x}}&=\underset{{x\to 0}}{\mathop{{\lim }}}\frac{{4\cos 4x}}{3}\\&=\frac{{4\cos \left( 0 \right)}}{3}=\frac{{4\cdot 1}}{3}=\frac{4}{3}\end{align}$

$ \displaystyle \underset{{x\to -\infty }}{\mathop{{\lim }}}\,\frac{x}{{{{x}^{2}}-1}}$

Indeterminate Form: $ \displaystyle \frac{0}{0}$

Divide each term by $ {{x}^{2}}$, since this is the highest degree (exponent). Remember that $ \displaystyle \underset{{x\to -\infty }}{\mathop{{\lim }}}\,\frac{1}{x}=0$:

$ \displaystyle \begin{align}\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\,\frac{x}{{{{x}^{2}}-1}}&=\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\,\frac{{\frac{x}{{{{x}^{2}}}}}}{{\frac{{{{x}^{2}}}}{{{{x}^{2}}}}-\frac{1}{{{{x}^{2}}}}}}\\&=\underset{{x\to \,-\infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{x}}}{{1-\frac{1}{{{{x}^{2}}}}}}=\frac{0}{{1-0}}=0\end{align}$

Take the derivative of the top and bottom of the fraction (do this twice!), and then take the limit:

$ \displaystyle \begin{align}\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\frac{x}{{{{x}^{2}}-1}}&=\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\frac{1}{{2x}}\\&=\underset{{x\to -\,\infty }}{\mathop{{\lim }}}\frac{0}{2}=0\end{align}$

Since there is no $ x$ in the answer, ignore $ x$ going to $ -\infty $.

$ \displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sqrt[3]{{h+27}}-3}}{h}$

Indeterminate Form: $ \displaystyle \frac{0}{0}$

Too complicated to do algebraically!

Remember that you don’t need the quotient rule; just take derivative on top and bottom:

$ \begin{align}\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sqrt[3]{{h+27}}-3}}{h}\,&=\,\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{{{{\left( {h+27} \right)}}^{{\frac{1}{3}}}}-3}}{h}=\,\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\frac{1}{3}{{{\left( {h+27} \right)}}^{{-\frac{2}{3}}}}-0}}{1}\\&=\,\frac{1}{3}{{\left( {0+27} \right)}^{{-\frac{2}{3}}}}=\frac{1}{3}{{\left( {\frac{1}{{27}}} \right)}^{{\frac{2}{3}}}}=\frac{1}{3}\left( {\frac{1}{9}} \right)=\frac{1}{{27}}\end{align}$

$ \displaystyle \underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {\pi +t} \right)}}{t}$

Indeterminate Form: $ \displaystyle \frac{0}{0}$

Not sure how to do this algebraically!

Take the derivative of the top and bottom of the fraction, and then take the limit:

$ \displaystyle \begin{align}\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\sin \left( {\pi +t} \right)}}{t}&=\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\cos \left( {\pi +t} \right)}}{1}\\&=\cos \left( {\pi +0} \right)=-1\end{align}$

Learn these rules, and practice, practice, practice!

Use the MathType keyboard to enter a Limit problem, and then click on Submit (the arrow to the right of the problem) to solve the problem using L’Hopital’s Rule. You can also click on the 3 dots in the upper right hand corner to drill down for example problems.

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On to Riemann Sums and Area by Limit Definition – you are ready!