Limit Question: What exactly happens in this two limits where $x \to \infty$ Hey guys so I have these two limits:
$$\lim_{x \to ∞} f(x) = \frac{1-e^x}{e^{2x}}=\frac{e^{-x}-1}{e^x}=\frac{1}{∞}=0$$
$$\lim_{x \to ∞} f(x) = \frac{\sqrt{x}}{1-\sqrt{x}}=\frac{1}{x^{-1/2}-1}=-1$$
As funny as it sounds I know that these limits are right; mechanically I know they're supposed to be like that but when I try to do them step by step; I can't get what the expressions have! any help step by step?
 A: For the first one, divide the numerator and denominator by  $e^x $ and for the second one,divide the numerator and denominator by  $\sqrt{x} $. Also note that $\frac{1}{a^b}=a^{-b} $.
A: It's not really different from what one does for limits at $\infty$ with polynomials and fractions
Consider $x^2-x$; written like this, it is an “indeterminate form” $\infty-\infty$, but it is sufficient to write it as
$$
x^2\left(1-\frac{1}{x}\right)
$$
to remove the “indetermination”: the term in parentheses has limit $1$.
What we did is “guessing the dominant term”, the term that's “much bigger” than the others when $x$ is large.
With fractions we still guess the dominant term in both the numerator and the denominator, so computing the limit becomes just comparing the two.
What's the dominant term in $e^x-1$? Of course it is $e^x$. In $e^{2x}$ there's no choice, so we write
$$
\lim_{x\to\infty}\frac{e^x-1}{e^{2x}}=
\lim_{x\to\infty}\frac{e^x}{e^{2x}}\frac{1-\dfrac{1}{e^x}}{1}=
\lim_{x\to\infty}\frac{1-e^{-x}}{e^x}=0
$$
The dominant term in the denominator “wins”.
Let's do the second one. The dominant term in the numerator is $\sqrt{x}$; in the denominator it is again $\sqrt{x}$, so we have
$$
\lim_{x\to\infty}\frac{\sqrt{x}}{1-\sqrt{x}}=
\lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x}}\frac{1}{\dfrac{1}{\sqrt{x}}-1}=-1
$$
There's a draw between the dominant terms.
Actually, these are polynomials (and rational functions) in disguise, because
$$
\frac{e^x-1}{e^{2x}}=\frac{t-1}{t^2}
$$
with the substitution $t=e^x$. Similarly for the second limit.
However, the idea of the “dominant” term (if it exists, of course) is helpful also in other cases: think to
$$
\lim_{x\to\infty}\frac{x-\sin x}{x+\sin x}
$$
We (rightly) guess that the dominant term in both terms is $x$:
$$
\lim_{x\to\infty}\frac{x-\sin x}{x+\sin x}=
\lim_{x\to\infty}\frac{x}{x}\frac{1-\dfrac{\sin x}{x}}{1+\dfrac{\sin x}{x}}=1
$$
