Example that you are tempted to use L'Hopital rule even if the limit does not exist

70 Views Asked by At

L'Hospital Rule says that

Let $s$ signify $a,a^+,a^-,\infty,-\infty$.Suppose $f$ and $g$ are differentiable for which the following limit exists:

$$\lim_{x\to s}\frac{f'(x)}{g'(x)} = L \cdots\cdots\text{(1)}$$

If

$$\lim_{x\to s} f(x) = \lim_{x \to s} g(x) = 0 \cdots\cdots\text{(2)}$$

or if

$$ \lim_{x\to s} |g(x)| = +\infty \cdots\cdots\text{(3)}$$

then

$$\lim_{x\to s}\frac{f(x)}{g(x)} = L$$

I would like to know that, is there any example where people erroneously use L'Hospital Rule due to ignoring the assumptions?

For example, we might be tempted to do

$$\lim_{x\to 0}x\log(x)=\lim_{x\to 0}\frac{\log(x)}{\frac{1}{x}}=^{L'Hospital}\lim_{x\to 0} \frac{\frac{1}{x}}{-\frac{1}{x^2}}$$

even though this is wrong since $\log(x)$ is not defined on $(-\epsilon,\epsilon)$

1

There are 1 best solutions below

0
On BEST ANSWER

Here's an example of how things can go wrong if ($1$) is not satisified: $$ \lim_{x\to\infty}\frac{x+\sin x}{x}=1+\lim_{x\to\infty}\frac{\sin x}{x}=1 $$ but if we try to apply L'Hopital's rule we get $$ \lim_{x\to\infty}[1+\cos x] $$ which does not exist.