0
$\begingroup$

Consider the limit as x approaches 0 of $x^2 / \sin x$. Intuivitely, that should be zero, because $x^2$ goes to zero much faster than $x$ does, and, near zero, $\sin x$ is approximated by $x$.

That intuition is captured by L'Hopital's rule. So, I would use L'Hopital's rule, rewriting it to $x/(\sin x / x)$, which, taking the derivative, is $1 / (\cos x / x - \sin x / x^2)$, which is still undefined for $x = 0$.

This confuses me for two reasons:

  1. If correct, shouldn't that mean there is no limit ($1/0$), not a limit of 0?
  2. When L'Hopital's rule is "stuck", how do you advance the problem?
  • 2
    To answer your first question, note that L'Hopital's doesn't apply to $x/(\sin x/x)$, which is of the form $0/\text{(indeterminate)}$. You need to resolve the indeterminate portion before you can proceed with this approach. Once you resolve the indeterminate form to $1$, you'll find the limit in question is zero. – 2017-02-09

3 Answers 3

1

Q1: The limit is not of $1/0$ form. Notice that $\lim_{x\to0}\frac1{\frac{\cos x}x-\frac{\sin x}{x^2}}=\frac1{\frac10-\frac00}$.

Q2: When L'Hospital's rule doesn't work, you should try rearrangement of the fractions. As many have mentioned, applying L'Hospital's rule to the original fraction works just fine.

Another approach would be to use series expansions, though highly unnecessary.

$$\sin x=x-\frac16x^3+\mathcal O(x^5)$$

$$\frac{x^2}{\sin x}=\frac x{1-\frac16x^2+\mathcal O(x^4)}\to\frac01=0$$

5

Why not just apply L'Hopital's rule to the original quotient? The result is $$ \lim_{x\to 0}\frac{x^2}{\sin x}=\lim_{x\to 0}\frac{2x}{\cos x}=0 $$

Alternately, you can use $\lim_{x\to 0}\frac{\sin x}{x}=1$ to obtain $$ \lim_{x\to 0}\frac{x^2}{\sin x}=\Big(\lim_{x\to 0}x\Big)\Big(\lim_{x\to 0}\frac{x}{\sin x}\Big)=0\cdot 1=0$$

2

There are two problems here. First is that you can't apply L'Hospital's Rule on $$\frac{x}{(\sin x)/x}$$ because the denominator $(\sin x)/x$ does not tend to $0$. Second is that one can easily apply L'Hospital directly on $x^{2}/\sin x$ to get $2x/\cos x \to 0$ as $x \to 0$.

Before applying L'Hospital's Rule one must check the conditions for its applicability (BTW most beginners don't follow this advice and just differentiate and plug to get rid of the "damn" limit problem).

Even after checking the conditions of applicability of L'Hospital's Rule on $f(x)/g(x)$ it is possible that the resulting expression $f'(x)/g'(x)$ does not have a limit. Or even worse the expression $f'(x)/g'(x)$ is complicated so that you don't know if this has a limit or not and you don't want to apply the rule second time because it may lead to even more complicated function $f''(x)/g''(x)$.

It is never a good idea to apply this rule directly (except for simplest limit problems). It is better to simplify the expression $f(x)/g(x)$ by algebraic manipulation and the use of standard limits and then apply the rule so that the expression obtained after applying the rule is simpler as far as limit evaluation is concerned.

Also you may get a problem where the ratio $f'(x)/g'(x)$ does not have a limit and yet $f(x)/g(x)$ has a limit. In such cases the rule is useless (but you can't know this before applying the rule) and then you need other methods like standard limits, Squeeze theorem, Taylor's series.