Confusion: When can I preform operation of infinity in limit (without using the explanation of Epsilon Delta Definition) Question 1:
$$\lim_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}$$
Knowing this:
$$\lim_{x\to\infty}\frac{1}{x}$$
The denominator of question 1 is $\infty$, therefore, $\lim_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}} = 0$
My confusion: Why could we assume that $\sqrt{\infty+1}+\sqrt{\infty} = \infty$ since there aren't any algebraic operation on inf and negative inf? e.g. This operation is wrong $\lim_{x\to\infty}x-x = \infty - \infty$
Question 2:
$$\lim_{x\to 0^{-}}\frac{1+2^{\frac{1}{x}}}{3+2^{\frac{1}{x}}}$$
Knowing that $\lim_{x\to 0^{-}}\frac{1}{x}$ and $\frac{1+2^{\frac{1}{x}}}{3+2^{\frac{1}{x}}}$ is continuous.

Using this theorem

$$\frac{1+2^{\lim_{x\to 0^{-}}\frac{1}{x}}}{3+2^{\lim_{x\to 0^{-}}\frac{1}{x}}}$$
$$\frac{1+2^{-\infty}}{3+2^{-\infty}}$$
since $2^{-\infty} = \frac{1}{2^{\infty}}$, Therefore
$$\frac{1+2^{-\infty}}{3+2^{-\infty}} = \frac{1}{3}$$
My confusion: Why could we assume that $2^{-\infty} = \frac{1}{2^{\infty}}$?
 A: I usually substitute $-\frac{1}{M}$ for $x$, in case $x\rightarrow 0^{-}$ and $\frac{1}{M}$ in case $x\rightarrow 0^{+}$ and then let $M\rightarrow\infty$. For example,
$$
\lim_{x\rightarrow 0^{-}}\frac{1+2^{\frac{1}{x}}}{3+2^{\frac{1}{x}}}=\lim_{M\rightarrow\infty}\frac{1+2^{-M}}{3+2^{-M}}=\lim_{M\rightarrow\infty}\frac{2^{M}+1}{3.2^{M}+1}=\lim_{y\rightarrow\infty}\frac{y+1}{3y+1}.
$$
It reduced to a rational function limit problem.
A: Question 1:
Note that in your case both $\sqrt{1+x}$ and $\sqrt{x}$ have the same sign, i.e., you do not get anything of the type $\infty-\infty$.
Question 2:
Note that for every $\epsilon>0$ there exists some $\delta>0$ such that if $-\delta<x<0$ then $2^{\frac{1}{x}}<\epsilon=2^{\frac{1}{\delta}}$.
A: If you are confused about this, then don't write equations with $\infty$ in there.  Instead write in terms of limits:
As $x \to \infty$ we have:
$$
\sqrt{x} \to \infty\\
\sqrt{x+1} \to \infty\\
\sqrt{x}+\sqrt{x+1} \to \infty\\
\frac{1}{\sqrt{x}+\sqrt{x+1}} \to 0 .
$$
And: as $x \to 0^-$ we have
$$
\frac{1}{x} \to -\infty\\
2^{1/x} \to 0\\
1+2^{1/x} \to 1\\
3+2^{1/x} \to 3\\
\frac{1+2^{1/x}}{3+2^{1/x}} \to \frac{1}{3}
$$
If you write it out in steps like this, then maybe you will find out which of the steps you do not understand.
Only after you have lots of experience doing it this multi-step way should you start writing equations with $\infty$ in there; but always keep in mind that what it really means is the multi-step calculations.  With that "lots of experience" you can do it in your head.
A: For question 2:
Let $g(x) = 2^{\frac{1}{x}} and f(x) = \frac{1+x}{3+x}$
Using the theorem
$$lim_{x\to0^-}f(g(x))$$
$$=lim_{x\to0^-}\frac{1+2^{\frac{1}{x}}}{3+2^{\frac{1}{x}}}$$
$$=\frac{1+lim_{x\to0^-}2^{\frac{1}{x}}}{3+lim_{x\to0^-}2^{\frac{1}{x}}}$$
$$=\frac{1+lim_{x\to0^-}\frac{1}{2^{-\frac{1}{x}}}}{3+lim_{x\to0^-}\frac{1}{2^{-\frac{1}{x}}}}$$
Since $lim_{x\to\infty}2^{x} = \infty$
$$\frac{1+0}{3+0}=\frac{1}{3}$$
