It is ok. And I meant LHL and RHL of a limit. U know a limit has two sides

@Servaes I think he means the right and left/from above and from below limits ($x>0$ and $x<0$).

The right limit exists, and as you can see you can compute it if you transform the expression accordingly. When you get $\infty/\infty$ that doesn't tell you much, thus you need to transform/bound the numerator and denominator. Consider a simpler example: $\lim_{n\rightarrow \infty}{n/n^2}$ you could say that's $\infty/\infty$ too, but if you divide you get $0$.

A limit of the form $\infty/\infty$ does not allow you to conclude that it does not exist. It simply means more effort is needed to determine whether it exists. Notice that very easy-to-understand limits like $\lim_{x\to \infty} x/x$ fit the same pattern.

$\frac \infty\infty$ is called "indeterminate" (as is $\frac 00$ and $0\cdot \infty$) in that it is ... not determinable. It might exist and it might be anything or it might not exist. Your transforming was enough to figure it out.

How do you know $\lim_{h \to 0}\frac{1 - \frac{1}{e^{1/h}}}{1 + \frac{1}{e^{1/h}}} = 1$ ?

@SeleneAuckland by substituting h=0, you get 1

@SeleneAuckland $1/0 = \infty, e^{\infty} = \infty, 1/\infty = 0$

@rash Why is $1/0 = \infty$ when $\lim_{x \to 0^{-}} \frac1x = -\infty$ ?

@SeleneAuckland What u said is for $\lim\limits_{x\to 0^-}\frac{1}{x}$, not for $\lim\limits_{h\to 0}\frac{1}{h}$. The limit with $x \to 0^-$ is converted to $h \to 0$.

@rash What do you mean by "converted" ? See wolfram.

@SeleneAuckland check the chat

@rash See what your accepted answerer has to say.

@rash "@SeleneAuckland to make calculation of limits easier x \to 0 can be converted to h \to 0 " What is "converted" ?

@SeleneAuckland $x - 0^- = h$,$h = x+0^-$ so $h \to 0$.

@SeleneAuckland what is your age?