Limit without L'Hopital rule How to compute the limits for the following functions without using L'Hopital rule? 
1) $\displaystyle\underset{x\to 0^{+}}{\lim} \frac{e^{1/x} + 2 e ^{-1/x} + \ln x}{3e^{1/x} + 5e^{-1/x} + 7\ln x}  $
2) $\displaystyle\underset{x\to 0^{+}}{\lim} \frac{x+e^{-1/x}}{x-e^{-1/x}}$
I wouldn't know where to begin. Thank you very much. 
 A: For problem 1, I did this: 
Take out $e^{1/x}$
$$\lim_{x\to 0} \frac{1+ 2\,e^{-2/x}+e^{-1/x}\ln x}{3 + 5\,e^{-2/x} + 7\,e^{-1/x}\ln x}$$
Now what? 
For problem 2, I did this: 
Substitute $y=1/x$
$$\lim_{x\to 0} \frac{1/y + e^{-y}}{1/y - e^{-y}} $$
Now what? 
Thank you very much. 
A: Following the hint for 1) you get
$$
\require{cancel}
\lim_{x\to 0^{+}} \frac{\cancel{e^{1/x}}}{\cancel{e^{1/x}}}\frac{1 + 2e^{-2/x} + e^{-1/x}\ln x}{3 + 5e^{-2/x} + 7e^{-1/x}\ln x}
$$
And then we observe that, for $x\to 0$, $e^{-1/x}$ wins over $\ln x$
$$
\left\vert \frac{\ln x}{e^{1/x}} \right\vert \leq \left\vert \frac{x-1}{e^{1/x}} \right\vert \underset{x\to0}{\longrightarrow} 0
$$
And I used a well-known logarithm inequality. So the limit is evaluates to
$$
\lim_{x\to 0^{+}} \frac{1 + 2e^{-2/x} + e^{-1/x}\ln x}{3 + 5e^{-2/x} + 7e^{-1/x}\ln x} = \frac{1}{3}
$$
Because $e^{-2/x}\to 0$ for $x\to 0$.
For problem 2), following from your answer, do the same trick: get $y$ out of the fraction, and then 
$$
\frac{e^{-y}}{y}\underset{y\to\infty}{\longrightarrow} 0
$$
(This is well-known, no inequalities involved) So that the limit for 2) is 1.
