evaluating some limits - calculus Problems with the following limits:
$$
1. \quad \quad  \lim_{x\to0^+} e^{1/x} + \ln x \, .
$$
Substitutions such as $e^{1/x}=t$ and $1/x = t$ don't yield any useful results. 
Pretty much the same with
$$
2. \quad \quad \lim_{x\to 0^+} e^{1/x} - 1/x \, ,
$$
Common denominator doesn't help much. 
Thanks. 
 A: If you accept that $\ln u\lt u$ for all $u$, it follows that $\ln(1/x)\lt1/x$, hence
$$e^{1/x}-1/x\lt e^{1/x}-\ln(1/x)=e^{1/x}+\ln x$$
Consequently, it suffices to show
$$\lim_{x\to0^+}(e^{1/x}-1/x)=\infty$$
Now we certainly have $\lim_{x\to0^+}e^{1/x}=\infty$.  Let's write
$$e^{1/x}-1/x=e^{1/x}\left(1-{1/x\over e^{1/x}} \right)$$
and take a look at
$$\lim_{x\to0^+}\left(1-{1/x\over e^{1/x}} \right)=1-\lim_{x\to0^+}{1/x\over e^{1/x}}=1-\lim_{x\to0^+}{-1/x^2\over(-1/x^2)e^{1/x}}=1-\lim_{x\to0^+}{1\over e^{1/x}}=1-0=1$$
This feeds into the general theorem that if $\lim f(x)=\infty$ and $\lim g(x)\gt0$, then $\lim f(x)g(x)=\infty$.
A: For $2)$: $e^{1/x} - 1/x > 1 + 1/x^2 \to +\infty$, and for $1)$ $e^{1/x} + \ln x = e^{1/x} - \ln\left(1/x\right)> 1 + 1/x + 1/x^2 - \left(1/x - 1\right)> 1/x^2 \to +\infty$
A: \begin{align*}
\lim \limits _{x\rightarrow 0^+} e^{\frac{1}{x}} +\ln x& = \lim \limits _{x\rightarrow 0^+} \frac{1}{x} \frac{e^{\frac{1}{x}} +\ln x}{\frac{1}{x}}\\
&= \lim \limits _{x\rightarrow 0^+} \frac{1}{x} \lim \limits _{x\rightarrow 0^+} \frac{e^{\frac{1}{x}} +\ln x}{\frac{1}{x}}\\
&= \lim \limits _{x\rightarrow 0^+} \frac{1}{x} \lim \limits _{x\rightarrow 0^+} \frac{\frac{-1}{x^2}e^{\frac{1}{x}} + \frac{1}{x}}{\frac{-1}{x^2}}\\
&= \lim \limits _{x\rightarrow 0^+} \frac{1}{x} \lim \limits _{x\rightarrow 0^+} e^{\frac{1}{x}}  -x\\
&=+\infty
\end{align*}
The second one is like the first one.
A: first let us show that $$  \lim_{u \to \infty} \frac{\ln u}{e^u} = 0$$ this follows because $$ \ln u < u < \frac{u^2}{2} < e^u \text{ so that } \frac{\ln u}{e^u} < \frac{2u}{u^2} \rightarrow 0 \text{ as } u \to \infty. $$
let us make a change of variable $u = \frac 1 x, x = \frac 1 u.$ with that we nee to find the $$\lim_{u\to \infty} e^u - \ln u = \lim_{u \to \infty}e^u\left(1- \frac{\ln u}{e^u}\right) =\infty \times 1 = \infty. $$
