Question regarding $\lim_{x \to 0} \left(\exp(\sin (x)) + \exp \left(\frac{1}{\sin (x)}\right)\right)$

Question

I wanted to find out whether the following limit exists, and find the value if it does. $$\lim_{x \to 0} \left(\exp(\sin (x)) + \exp \left(\frac{1}{\sin (x)}\right)\right).$$

Attempt
After many attempt to prove that the limit exists, I looked up Wolfram Alpha, and it turns out that it doesn't. So I set out to prove that. By the algebra of limits we have $$\lim_{x \to 0} \left(\exp(\sin (x)) + \exp \left(\frac{1}{\sin (x)}\right)\right)= \lim_{x \to 0} \exp(\sin (x)) + \lim_{x \to 0} \exp \left(\frac{1}{\sin (x)}\right).$$

Now, let $$f(x)=\frac{1}{\sin (x)} \ \text{and} \ g(x)=\exp (x),$$ and consider the sequences $$(y_n)=\dfrac{1}{\frac{\pi}{2}+ 2n\pi} \ \text{and} \ (z_n)=\dfrac{1}{-\frac{\pi}{2}+ 2n\pi }\text{.}$$

Then $(y_n) \rightarrow 0$ and $(z_n) \rightarrow 0$ as $n \rightarrow 0$, however $f(y_n)=1$ and $f(z_n)=-1$ for all $n \in \mathbb{N}$.

Hence $\lim_{n \to \infty}f(y_n) \neq \lim_{n \to \infty}f(z_n)$ and $\lim_{x \to 0} \frac{1}{\sin (x)}$ does not exist.

If the working above is correct, I then would like to show that

1. since $\lim_{x \to 0} \frac{1}{\sin (x)}$, then $\lim_{x \to 0} \exp \left(\frac{1}{\sin (x)}\right)$ does not exist,

2. if $\lim_{x \to 0} \exp \left(\frac{1}{\sin (x)}\right)$ does not exist, then $\lim_{x \to 0} \exp(\sin (x)) + \lim_{x \to 0} \exp \left(\frac{1}{\sin (x)}\right)$ does not exist.

Question
A. Is my working above correct?

B. How do I show (1) and (2)? I know the rule for finding the limit of the composition of two functions and the addition of limits, if the limit exists, but not if the limit doesn't exist.

C. Is there any other way to tackle this question?

D. Could you give me some tips to identify if the limit of a function does not exist, without any aid of computers? (In order to not waste time trying to prove the existence of the limit when it actually doesn't.)

Thank you for your time.

Answer 1

Note that $\frac{1}{\sin x}$ blows up as $x$ approaches $0$ from the right. So the limit does not exist. Even if we allow infinite limits, the limit does not exist, for our function approaches $1$ as $x$ approaches $0$ from the left.

Remark: If the question is changed to $\lim_{x\to 0^-}$ (limit as $x$ approaches $0$ from the left), then the limit does exist, and is equal to $1$.

The calculations in the post for values of $x$ of special form are not useful for this problem. The function $\sin x$ behaves very smoothly for $x$ near $0$, and it makes no difference if $x$ is of special shape such as $\frac{1}{2n\pi+\frac{\pi}{2}}$. Such values of $x$ would be very important if we were studying expressions that involve $\sin(1/x)$ for $x$ near $0$.

As to your question about ways other than using computers, it is a matter of visualizing the behaviour of these fairly familiar functions. For positive $x$ near $0$, $\sin x$ is close to $0$ and positive, so $\exp(1/\sin x)$ is really really big. For negative $x$ near $0$, $\frac{1}{\sin x}$ is big negative, so $\exp(1/\sin x)$ is close to $0$.