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L’Hopital’s Rule

In Derivative Using the Limit Definition, we saw that derivatives were defined using limits. It turns out, that after we take the time to find the derivatives of certain functions, derivative can indeed be used to find limits that we will run into. In fact, using what is called L’Hopital’s rule, we have a nice way to get at many different limits that, in the past, have required many different tricks to get to.

L’Hopital’s rule

Linear approximations of functions

Before we get into using L’Hopital’s rule, we should take some time to cover what it says and why it works. The first thing we need to remember is the limit definition of derivative, in particular, at a given point \(x_{0}\), we have that
\begin{align*}
f'(x_{0})=\lim_{\Delta x \to 0}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}.
\end{align*}
Now, since the limit tells us what happens as \(\Delta x\) get extremely close to \(0\), we note that, for small \(\Delta x\)s, we have that
\begin{align*}
f'(x_{0}) \approx \frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}.
\end{align*}
That is, we can get a good approximation for the derivative is we use small changes in \(x\) and we can make this exact by taking a limit.

If we rearrange this approximation, we can get that
\begin{align*}
f'(x_{0})\Delta x & \approx f(x_{0}+\Delta x)-f(x_{0}).
\end{align*}
Furthermore, if we let \(z=x_{0}+\Delta x\), we get that
\begin{align*}
f(z) & \approx f'(x_{0})(z-x_{0})+f(x_{0}).
\end{align*}
That is, our function is approximately the same as a linear function. Furthermore, this linear function will be a good approximation for small changes in \(x\) and will become exact as we let the change in \(x\) get arbitrarily small.

L’Hopital’s rule

Now, suppose that \(f(x)\) and \(g(x)\) are differentiable functions and that we want to find
\begin{align*}
\lim_{x \to c}\frac{f(x)}{g(x)}.
\end{align*}

The first thing we should do is see what happens at \(x=c\). Note that, since \(f\) and \(g\) are differentiable, \(f(c)\) and \(g(c)\) exist. If \(g(c) \neq 0\), then \(\frac{f(c)}{g(c)}\) is a number, so the limit will just be this number. Furthermore, if \(f(c)\) is a non-zero number and \(g(c)=0\), then we have seen that we will have a vertical asymptote.

The only case we have left to worry about is the case where \(g(c)=f(c)=0\). In this case, we would have that
\begin{align*}
\lim_{x \to c}\frac{f(x)}{g(x)}&=\lim_{x \to c}\frac{f'(c)(x-c)+f(c)}{g'(c)(x-c)+g(c)},
\end{align*}
using the linear approximations from above. Furthermore, since \(x\) is getting arbitrarily close to \(c)\) we get that these are not only approximation, but exact. Then, since \(f(c)=g(c)=0\), we get
\begin{align*}
\lim_{x \to c}\frac{f(x)}{g(x)}&=\lim_{x \to c}\frac{f'(c)(x-c)+f(c)}{g'(c)(x-c)+g(c)}, \\
&=\lim_{x \to c}\frac{f'(c)(x-c)+0}{g'(c)(x-c)+0} \\
&=\lim_{x \to c}\frac{f'(c)(x-c)}{g'(c)(x-c)} \\
&=\frac{f'(c)}{g'(c)},
\end{align*}
provided that \(g'(c) \neq 0\). If \(g'(c)=0\), then we run into a situation where the original limit may still exist. In order to find if it does, we would have to find an approximation with the 2nd, 3rd, 4th or… derivative. If we do address this issue, we will find that
\begin{align*}
\lim_{x \to c}\frac{f(x)}{g(x)}&=\lim_{x \to c}\frac{f'(x)}{g'(x)}.
\end{align*}

Putting this together

While we may want to be more careful with an exact proof, the idea is that we can use our linear approximations using derivatives. That is, if we now run into the indeterminate form of limits \(\frac{0}{0}\), L’Hopital’s rule gives us a method that will allow us to find the limit without having to resort to different algebraic tricks. We can now just find the limit of the derivative of the top divided by the derivative of the bottom.

Without going into more detail, we note that we can also use L’Hopital’s rule if we have limits of the indeterminate forms \(\frac{\pm \infty}{\pm \infty}\) and if we are taking the limit as \(x \to \pm \infty\).

Examples

Now that we know what L’Hopital’s rule says, we are ready to use it.

First example

Find
\begin{align*}
\lim_{x\to 1}\frac{\ln(x)}{x-1}.
\end{align*}

Note that in this case, we have that \(\ln(1)=1\) and \(1-1=0\). Therefore, this limit is of the form \(\frac{0}{0}\), so we can use L’Hopital’s rule. Since \(\ln(x)’=\frac{1}{x}\) and \((x-1)’=1\), we now get
\begin{align*}
\lim_{x\to 1}\frac{\ln(x)}{x-1}&=\lim_{x \to 1}\frac{\frac{1}{x}}{1}.
\end{align*}
We now have a new limit to calculate, which we hope will be easier to find. At this point, it is extremely helpful to simplify any of your limits if you can. Doing so, we see
\begin{align*}
\lim_{x\to 1}\frac{\ln(x)}{x-1}&=\lim_{x \to 1}\frac{\frac{1}{x}}{1} \\
&=\lim_{x \to 1}\frac{1}{x} \\
&=\frac{1}{1}=1.
\end{align*}
After simplifying, we were able to evaluate the funciton inside the limit at \(x=1\). Since we got an number out, this is our limit.

Using L’Hopital’s rule twice

Find
\begin{align*}
\lim_{x \to 0}\frac{\cos(x)-1}{x^{2}}.
\end{align*}

In this case, we again evaluate the inside function at \(x=0\). We then get \(\frac{\cos(0)-1}{0}=\frac{0}{0}\). We can therefore us L’Hopital’s rule. We then get
\begin{align*}
\lim_{x \to 0}\frac{\cos(x)-1}{x^{2}}&=\lim_{x \to 0}\frac{-\sin(x)}{2x}.
\end{align*}
We now have a new limit. If we evaluate this at \(0\), we get \(\frac{0}{0}\). Therefore, in order to find our new limit, we can again use L’Hopital’s rule. We therefore get
\begin{align*}
\lim_{x \to 0}\frac{\cos(x)-1}{x}&=\lim_{x \to 0}\frac{-\sin(x)}{2x} \\
&=\lim_{x \to 0}\frac{-\cos(x)}{2}.
\end{align*}
We have, again, arrived at a new limit. We then evaluate the function inside the limit at \(0\) and get \(\frac{-1}{2}\). Finally, we get a number out, so this is our limit. We now have that
\begin{align*}
\lim_{x \to 0}\frac{\cos(x)-1}{x}&=\lim_{x \to 0}\frac{-\sin(x)}{2x} \\
&=\lim_{x \to 0}\frac{-\cos(x)}{2} \\
&=\frac{-1}{2}.
\end{align*}

Converting a product to a quotient

Find
\begin{align*}
\lim_{x \to \infty}x\ln(1+\frac{2}{x}).
\end{align*}

Here, we will again determine what happens to the function inside the limit as \(x \to \infty\). In this case we will have something that looks like \(\infty*\ln(1+0)=\infty * 0\). Now, \(\infty*0\) is also an indeterminate limit; however, it is not of the form \(\frac{0}{0}\) or \(\frac{\pm \infty}{\pm \infty}\), so we can not use L’Hopital’s rule. If we could convert it into one of forms, we could use L’Hopital’s rule. Now, if we want to convert a product into a quotient, we will just need to divide by the reciprocal of one of the functions. While you can use either function, this seems to look easier if we move the \(x\) to the bottom. That is, we will divide by \(\frac{1}{x}\), which is the same as multiplying by \(x\). We then get
\begin{align*}
\lim_{x \to \infty}x\ln(1+\frac{2}{x})&=\lim_{x \to \infty}\frac{\ln(1+\frac{2}{x})}{\frac{1}{x}}.
\end{align*}
If we now evaluate this function as \(x \to \infty\), we will get \(\frac{0}{0}\). Now, we use a chain rule to take the derivative of the top. Here we have \(o(i)=\ln(i)\), \(i=1+\frac{2}{x}\), \(o’=\frac{1}{i}\) and \(i’=\frac{-2}{x^{2}}\). We now have
\begin{align*}
\lim_{x \to \infty}x\ln(1+\frac{2}{x})&=\lim_{x \to \infty}\frac{\ln(1+\frac{2}{x})}{\frac{1}{x}} \\
&=\lim_{x \to \infty}\frac{\frac{1}{1+\frac{2}{x}}\frac{-2}{x^{2}}}{\frac{-1}{x^{2}}}.
\end{align*}

As we did before, we want to simplify our new limit before evaluating it. We then get
\begin{align*}
\lim_{x \to \infty}x\ln(1+\frac{2}{x})&=\lim_{x \to \infty}\frac{\ln(1+\frac{2}{x})}{\frac{1}{x}} \\
&\lim_{x \to \infty}\frac{\frac{1}{1+\frac{2}{x}}\frac{-2}{x^{2}}}{\frac{-1}{x^{2}}}\\
&=\lim_{x \to \infty}\frac{2}{1+\frac{2}{x}}.
\end{align*}
Now that we’ve simplified the problem, we can find that the function will go to \(\frac{2}{1+0}\) as \(x \to \infty\). Since this is a number, \(2\), this is our limit. Therefore,
\begin{align*}
\lim_{x \to \infty}x\ln(1+\frac{2}{x})&=\lim_{x \to \infty}\frac{\ln(1+\frac{2}{x})}{\frac{1}{x}} \\
&\lim_{x \to \infty}\frac{\frac{1}{1+\frac{2}{x}}\frac{-2}{x^{2}}}{\frac{-1}{x^{2}}}\\
&=\lim_{x \to \infty}\frac{2}{1+\frac{2}{x}}\\
&=2.
\end{align*}

Turning an exponent into a product

Find
\begin{align*}
\lim_{x \to 0^{+}}x^{x}.
\end{align*}

For this limit, we evaluate at \(0\) and get \(0^{0}\). Again, this is another indeterminate form for limits that is not one we can use L’Hopital’s rule. The issue here is that we have an exponent, but we want a quotient. However, if we could rewrite this as a product, we already know how to turn a product into a quotient. We, therefore, just have to figure out how to turn an exponent into a product. This is precisely what we can do using logarithm. Therefore, we want to take the natural log of the function. Since we don’t want to change the problem, we will also need the exponential function. We now get
\begin{align*}
\lim_{x \to 0^{+}}x^{x}&=e^{\ln(\lim\limits_{x \to 0^{+}}x^{x})}.
\end{align*}
Since \(\ln(x)\) and \(e^{x}\) are continuous are their domains, we may pull the limit to the outside of the \(\ln()\) as long as the inside limit is in the domain of \(\ln(x)\). We now have,
\begin{align*}
\lim_{x \to 0^{+}}x^{x}&=e^{\ln(\lim\limits_{x \to 0^{+}}x^{x})} \\
&=e^{\lim\limits_{x \to 0^{+}}\ln(x^{x})} \\
&=e^{\lim\limits_{x \to 0^{+}}x\ln(x)}.
\end{align*}
In the last step, we use logorithms properties to pull the exponent out as a multiple. In order to help keep us from making mistakes we will now focus on
\begin{align*}
\lim\limits_{x \to 0^{+}}x \ln(x).
\end{align*}
Here, we get \(0*-\infty\), so we will need to convert the product into a quotient. We then get
\begin{align*}
\lim\limits_{x \to 0^{+}}x \ln(x) &=\lim\limits_{x \to 0^{+}}\frac{\ln(x)}{\frac{1}{x}} \\
&=\lim\limits_{x \to 0^{+}}\frac{\frac{1}{x}}{\frac{-1}{x^{2}}}.
\end{align*}
We used L’Hopital’s rule to find the second step. We can then simplify and find
\begin{align*}
\lim\limits_{x \to 0^{+}}x \ln(x) &=\lim\limits_{x \to 0^{+}}\frac{\ln(x)}{\frac{1}{x}} \\
&=\lim\limits_{x \to 0^{+}}\frac{\frac{1}{x}}{\frac{-1}{x^{2}}} \\
&=\lim\limits_{x \to 0^{+}}-x \\
&=0.
\end{align*}

Now that we’ve found this limit, we are ready to find our original limit. We now have
\begin{align*}
\lim_{x \to 0^{+}}x^{x}&=e^{\ln(\lim\limits_{x \to 0^{+}}x^{x})} \\
&=e^{\lim\limits_{x \to 0^{+}}\ln(x^{x})} \\
&=e^{\lim\limits_{x \to 0^{+}}x\ln(x)} \\
&=e^{0} \\
&=1.
\end{align*}

Conclusion

We have seen now that are able to find limits that we weren’t previously able to find. In the case we have indeterminate forms, we were able to use L’Hopital’s rule instead of having to algebraically simplify the problem. This is an extremely useful method for finding limits much quicker than we otherwise would be able to.

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