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.
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 $\lim f(x)g(x)=\infty$.
first let us show that $$  \lim_{u \to \infty} \frac{\ln u}{e^u} = 0$$ this follows because $$ \ln u &lt; u &lt; \frac{u^2}{2} &lt; e^u \text{ so that } \frac{\ln u}{e^u} &lt; \frac{2u}{u^2} \rightarrow 0 \text{ as } u \to \infty. $$
let us make a change of variable $u = \frac 1 x, x = \frac 1 u.$ with that we nee to find the $$\lim_{u\to \infty} e^u - \ln u = \lim_{u \to \infty}e^u\left(1- \frac{\ln u}{e^u}\right) =\infty \times 1 = \infty. $$