Sep 29

Using l’Hôpital’s Rule

By Tutor GuyNo Comments

 

There are two guidelines you should always follow when applying l’Hôpital’s rule:

  • Make sure the original function gives you an indeterminate result before you take the derivatives;
  • Always simplify the result before plugging in to save yourself some extra work.

Example 1: Here’s a simple example that demonstrates the importance of the first guideline.

Find:

\displaystyle \lim_{x \to 0} \frac{x}{1+ \sin x}

 Solution: This function is continuous at x=0. To find the limit, simply plug in 0:

 \displaystyle \lim_{x \to 0} \frac{x}{1+ \sin x}=\frac{0}{1+0}=0

This is easily verified if you graph the function. However, if you try to apply l’Hôpital’s rule right away, you will get the incorrect value for the limit:

\displaystyle \lim_{x \to 0} \frac{x}{1+ \sin x}=\lim_{x \to 0} \frac{1}{\cos x}= \frac{1}{1}=1

 

Example 2: This example demonstrates the importance of the second guideline.

Find:

\displaystyle \lim_{x \to 0^+} \dfrac{\ln x}{^1 \!/ \!_x}

This satisfies the conditions for applying l’Hôpital’s rule, because plugging in gives an indeterminate form.

 

\displaystyle \lim_{x \to 0^+} \dfrac{\ln x}{^1 \!/ \!_x}=\frac{\ln 0}{^1 \! / _0}= \frac{-\infty}{\infty}

 We apply l’Hôpital’s rule once:

\displaystyle \lim_{x \to 0^+} \dfrac{\ln x}{^1 \!/ \!_x}= \lim_{x \to 0^+}\frac{1/x}{-1/x^2}

This would give us another indeterminate form if we plug in now, but we remember to simplify first, and it’s easy to evaluate:

 

\displaystyle \lim_{x \to 0^+} \dfrac{\ln x}{^1 \!/ \!_x}= \lim_{x \to 0^+}\frac{1/x}{-1/x^2}= \lim_{x \to 0^+} -x=0

Again, this is easy to verify by plotting the function.

Calculus
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