Can we use l-hospital to prove whether limit exists or not?

Question

My instructor recently gave me a question to find whether the limit of $$\lim_{x\to 0} \frac {((\sin^2(3x)/x^2)-9)}{x} $$ exists or not. Some person used L-Hospital rule and differentiated thrice and found limit value equal to zero. My question is whether this is the correct method or not? I think that when you use L-Hospital you already assume the fact that limit exists and hence it is kind of a circular argument and therefore not a valid method. Another way i think i can ask this question is if f'(x)/G'(x) where x->0 exists, does that imply limit f(x)/g(x) exists by L-Hospital's rule?

Answer 1

It's the main point of L’Hôpital’s theorem ($0/0$ case):

Suppose the functions $f$ and $g$ are defined and differentiable in a punctured neighborhood of $c$, with $g'(x)\ne0$ in this punctured neighborhood and that $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0$. If $$ \lim_{x\to c}\frac{f'(x)}{g'(x)}=l $$ exists (finite or infinite), then also $$ \lim_{x\to c}\frac{f(x)}{g(x)}=l $$

You're possibly confused with one of the objections about the usage of L’Hôpital’s theorem, which is in connection to $$ \lim_{x\to 0}\frac{\sin x}{x} $$ because the knowledge of this limit is thought to be necessary in the computation of the derivative of the sine function. This objection can be dismissed in two ways:

1. a rigorous definition of the sine function is as a power series and, as such, its differentiability is not a problem and the limit mentioned above is an easy consequence;

2. once you have proved that the wheel is useful for some task, you don't need to repeat the proof every time you have to perform that task, you just use the wheel.

In this case the wheel is the derivative of the sine: once you have proved it is the cosine, you can forget how you proved this and simply use your knowledge.

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On the other hand, if you define $$ \operatorname{sinc}x=\begin{cases} \dfrac{\sin x}{x} & x\ne0 \\[6px] 1 & x=0 \end{cases} $$ then the function $\operatorname{sinc}$ is everywhere continuous and also differentiable, as it can be proved easily: $$ \operatorname{sinc}'x=\begin{cases} \dfrac{x\cos x-\sin x}{x^2} & x\ne0 \\[6px] 0 & x=0 \end{cases} $$ Then your limit is $$ \lim_{x\to0}\frac{9\operatorname{sinc}^23x-9}{x} $$ that is, the derivative of $9\operatorname{sinc}^23x$ at $0$. Use the derivative above and the chain rule to prove the given limit is $0$.

As you see, there's really no need to use L’Hôpital’s theorem in this case.

Answer 2

Yes, that is what L'Hopital's rule says: if $\lim_{x\to a}\dfrac{f'(x)}{g'(x)}$ exists, then $\lim_{x\to a}\dfrac{f(x)}{g(x)}$ exists too and they are equal. You don't need to assume first that the limit $\lim_{x\to a}\dfrac{f(x)}{g(x)}$ exists.

Answer 3

Yes of course it is allowed since

$$\frac{\sin^2(3x)}{x^2}=9\left(\frac{\sin(3x)}{3x}\right)^2\to 9$$

and

$$\frac {\frac{\sin^2(3x)}{x^2}-9}{x}$$

is an indeterminate form $\frac 0 0$.

As an alternative we have

$$\frac {\frac{\sin^2(3x)}{x^2}-9}{x}=\frac {\left(\frac{\sin (3x)}{x}+3\right)\left(\frac{\sin (3x)}{x}-3\right)}{x}=\left(\frac{\sin (3x)}{x}+3\right)\frac {\left(\frac{\sin (3x)}{x}-3\right)}{x} \to6\cdot 0=0$$

indeed

• $\frac{\sin (3x)}{x}+3=3\frac{\sin (3x)}{3x}+3 \to 6$
• $\frac{\frac{\sin (3x)}{x}-3}x=\frac1{x^2} \left(3x-\frac{27x^3}{6}+o(x^3)-3x\right)=-\frac{27x}{6}+o(x)\to 0$

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A more interesting limit would be

$$\lim_{x\to 0} \frac {((\sin^2(3x)/x^2)-9)}{\color{red}{x^2}}$$