$$\lim_{x \to 0} \bigg ({\frac{(1+x)^{\frac{2}{x}}}{e^2}}\bigg)^\frac{4}{\sin x}$$
I was solving this but I got the wrong answer. I want to figure out why it is wrong.
I did:
$$L=\lim_{x \rightarrow 0} \bigg ({\frac{(1+x)^{\frac{2}{x}}}{e^2}}\bigg)^\frac{4}{\sin x}$$ $$\ln L =\lim_{x \rightarrow 0} \bigg(\frac{4}{\sin x} \frac{2}{x} \ \ln(1+x) -\frac{4}{\sin x}\ln e^2\bigg)$$ $$\ln L =\lim_{x \rightarrow 0} \bigg(\frac{4}{\sin x} \ \frac{2}{x} \ \ln(1+x) -\frac{8}{\sin x}\bigg)$$
Since $\lim\limits_{x\to0} \frac{\ln(1+x)}{x}=1$,
$$\ln L =\lim_{x \rightarrow 0}\bigg(\frac{8}{\sin x}-\frac{8}{\sin x}\bigg)$$ $$\ln L=0$$ $$L=e^0 =1$$
However, the given answer is $e^{-4}$.
Can someone please tell me the mistake I made here. I would be grateful if someone shared the correct solution. Thanks.
See the post I made here for the mistake you made.
Essentially, suppose that $f$ and $g$ are two functions and you want to find $\lim_{x \to 0} f(x)-g(x)$. Suppose that you found two functions $h(x)$ and $j(x)$ such that $$ \lim_{x \to 0} \frac{f(x)}{h(x)} = 1, \lim_{x \to 0} \frac{g(x)}{j(x)} = 1 $$ You want to believe that $\lim_{x \to 0} f(x)-g(x) = \lim_{x \to 0} h(x)-j(x)$. Unfortunately, this is not true in general, and I explained why in the answer I link : the hidden terms.
Essentially, when you subtract two functions which have the same first order asymptotic near $0$, this exposes further terms of the Taylor series of that difference. Hence, those terms are important to the final answer.
In your case, for example, once you wrote $$ \lim_{x \to 0} \frac{4}{\sin x} \frac{2\ln(1+x)}{x} - \frac{8}{\sin x} $$ Take $f$ as the first function and $g$ as the second. You found that $\lim_{x \to 0} \frac{f(x)}{h(x)} = 1$ where $h(x) = \frac{8}{\sin x}$. Taking $j=g$ , you wished that $\lim_{x \to 0} f(x)-g(x) = \lim_{x \to 0} h(x)-j(x)$. That's exactly what your next step looked like : $\lim_{x \to 0} \frac{8}{\sin x} - \frac{8}{\sin x}$, which had limit $0$. So the problem was with the replacement, because the answer actually depends upon further terms in the Taylor expansions of $f$ and $g$.
So to answer this question, there is no way we can solve this question without using the Taylor series of some function. That's because first-order asymptotics are not good enough, as we saw with the failed attempt. Equivalently, one cannot avoid (effectively) the usage of the L'Hopital rule, because it is, to all intents and purposes, tantamount to the existence of Taylor series.
However, one should still know when one can make the change $f(x) \to h(x)$ and still get the same answer, and what function's Taylor series will be involved. For example, I'll start from your last correct step : $$ \lim_{x\to 0} \frac{4}{\sin x} \frac{2\ln(1+x)}{x} - \frac{8}{\sin x} $$ Factorizing the $\frac{8}{\sin x}$ out, we get $$ \lim_{x \to 0} \frac{8}{\sin x} \left(\frac{\ln(1+x)-x}{x}\right) $$ At this point, noticing that $\frac{x}{\sin x} \to 1$ as $x \to 0$, we can do $$ \lim_{x \to 0} 8\frac{x}{\sin x} \left(\frac{\ln(1+x)-x}{x^2}\right) $$ So we could, legitimately, replace the $\sin x$ by $x$ in the limit without changing the answer (because $\frac{x}{\sin x} \to 1$ as $x \to 0$). The advantage of doing this is clear once you observe the term in the bracket. As I mentioned earlier, one cannot avoid Taylor series in the solution of this problem (equivalently, one cannot avoid using the L'Hospital rule).
However, this approach makes it extremely clear as to which function's Taylor series is to be used : $\ln(1+x)$, near $x=0$. The Taylor series of this function reads $$ \ln(1+x) = x-\frac{x^2}{2}+o(x^2) $$ where $o(x^2)$ is a function such that $\lim_{x \to 0} \frac{o(x^2)}{x^2} = 0$, because it only contains higher powers of $x$ than $x^2$. Therefore, rearranging and using the definition of $o(x^2)$, $$ \lim_{x \to 0} \frac{\ln(1+x)-x}{x^2} = -\frac 12 $$ Substituting this back, the desired limit equals $8 \times -\frac 12 = -4$, and recalling that we took the logarithm long back, the answer to the question is $e^{-4}$.
The lesson to be learned here is the same as that of the linked post : if $\lim_{x \to 0} \frac{f(x)}{h(x)} =1$ and $\lim_{x \to 0} \frac{g(x)}{j(x)} =1$, it is NOT true that $\lim_{x \to 0}\frac{f(x)-g(x)}{h(x)-j(x)} = 1$ always.
In such a case, one has to locate the higher order terms in the difference $f(x)-g(x)$ and deal with them separately by Taylor expanding a suitable function to that order. That's what we did here, and is a fundamental skill in limit solving.