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On evaluating $$f(x)=\begin{cases} \frac{x\ln(\cos x)}{\ln(1+x^2)} & x\neq0,\\0& x=0\end{cases}$$ all the solutions I've seen used the L'Hôpital's rule to break it down; that way, you find out that the function is continuous at x=0, becoming 0 there.

I used $\lim_{x^2 \to 0} \frac{\ln(1+x^2)}{x^2}=0$ and logarithmic properties to simplify it to $\lim_{x \to 0}\ln(\cos x)^{ \frac{1}{x}}$;

Defining h as a small positive value, as x approaches 0 from the right, it becomes h. As cosx approaches 0 from the right, it becomes 1-h. Thus the right-hand limit becomes $\lim_{h \to 0^+}ln(1-h)^\frac{1}{h}$. This becomes $ln0=-\infty$

Similarly, as x approaches 0 from the left, it becomes -h. As cosx approaches 0 from the left, it becomes 1-h, the same as before. Thus the left-hand limit becomes $\lim_{h \to 0^+}ln(1-h)^\frac{-1}{h}$. This is $\lim_{h \to 0^+}ln\frac{1}{(1-h)^\frac{1}{h}}$, which becomes $ln\infty$, which is $\infty$.

As the limit doesn't seem to exist as per my evaluation, am I supposed to use only the L'Hôpital's Rule in indeterminate forms?

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  • $\begingroup$ You should explain your attempted solution in more detail; from what you've written I think a couple different things might have gone wrong, but I'm not exactly sure which. $\endgroup$ Commented Nov 18, 2020 at 12:05

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Your idea of using the fact that $\frac {\ln (1+t)} t \to 1$ as $ t \to 0$ is good so the question reduces to that of finding $\lim_{x \to 0} \frac {\ln (\cos x)} x$. But this last limit is $0$ by L'Hopital's Rule. I don't know how you concluded that this limit does not exist.

There are conditions to be satisfied for application of L'Hopital's Rule and a wrong application can lead to a wrong answer.

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  • $\begingroup$ I added my method to the question. Can you tell me where I messed up now? $\endgroup$
    – harry
    Commented Nov 19, 2020 at 1:56
  • $\begingroup$ Okay, I fixed it. In a limit, isn't cosx a number slightly less than 1, and not 1? And isn't 1/x infinity, as x approaches 0? $\endgroup$
    – harry
    Commented Nov 19, 2020 at 11:22
  • $\begingroup$ @HarryHolmes It is not true that $\cos h \sim 1-h$ for $h$ small. What is true is $\cos h \sim1-\frac {h^{2}} 2$. $\endgroup$ Commented Nov 19, 2020 at 11:32
  • $\begingroup$ @HarryHolmes In your analysis you are thinking that $f(h) \to 1$ through values less than $1$ as $h \to 0+$ implies $f(h)^{1/h}\to 0$. That is not correct. For example $(1-\frac 1 n)^{n} \to e^{-1}$. $\endgroup$ Commented Nov 19, 2020 at 11:38
  • $\begingroup$ Oh, right. Should I always use the expansions in trigonometric limits? $\endgroup$
    – harry
    Commented Nov 19, 2020 at 12:31
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" Am I supposed to use only the L'Hôpital's rule for indeterminate forms? "

No, not only LH rule for indeterminate forms.

For example, let's see the following limit: $$\lim_{n\to \infty} \frac{\ln n}{n}$$ (you can write $x$ instead of $n$, as well). Let's try to solve it without LH rule.

Note that $$0 \le \frac{\ln n}{n} = \frac{2\color{red}{\ln \sqrt n}}{\sqrt n \color{red}{\sqrt n}} \le \frac{2}{\sqrt n}\cdot \color{red}1 \\ 0 \le \frac{\ln n}{n} \le \frac{2}{\sqrt n}$$ Now, by Sandwich Theorem (or Squeeze Theorem), we conclude that $$\lim_{n\to \infty} \frac{\ln n}{n} = 0$$

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Hint:

We actually don't need The Intuition behind l'Hopitals Rule Or Series Expansion as we can write

$$2f(x)=-x\cdot\dfrac{\ln(1-\sin^2x)}{-\sin^2x}\cdot\dfrac{x^2}{\ln(1+x^2)}\cdot\left(\dfrac{\sin x}x\right)^2$$

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