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 $2^{-\infty} = \frac{1}{2^{\infty}}$?

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.
