<s>[INST] 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 $2^{-\infty} = \frac{1}{2^{\infty}}$?
 [/INST] 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 $lim_{x\to\infty}2^{x} = \infty$
$$\frac{1+0}{3+0}=\frac{1}{3}$$
</s>