How to do limit? Find the limit $$\lim_{x \to 0^+} (e^{\frac 6x}+8x)^{\frac x2}.$$
I tried taking the $\ln$ of both sides and bringing the exponent down, then moving the $x$ to the bottom as $x^{-1}$. Then the limit was $\infty/\infty$, so I could use l'Hopital's Rule. But that made the problem bigger (and a second use of l'Hop's made it even worse.
P.S. I can't figure out how to make that any more readable (I can't read it on my monitor). The exponents are $e$ raised to ($6$ over $x$)" and the outer exponent is "$x$ over $2$".
 A: You want
$\lim_{x \to 0^+} (e^{\frac 6x}+8x)^{\frac x2}
$.
The $e^{6/x} \to \infty$
and the
$8x \to 0$,
so this is the same as
$\lim_{x \to 0^+} (e^{\frac 6x})^{\frac x2}
$.
But
$(e^{\frac 6x})^{\frac x2}
=e^3
$
which is your answer.
A: I expand the comment of Andre Nicolas. Note that since $x > 0$ we have $$e^{6/x} > 1 + \frac{6}{x}$$ and $6/x \to \infty, 8x \to 0$ as $x \to 0^{+}$. Hence there is a $\delta > 0$ such that $6/x > 8x$ for all $x$ with $0 < x < \delta$. It now follows that $$e^{6/x} < e^{6/x} + 8x < 2e^{6/x}$$ for all $x$ with $0 < x < \delta$ and upon taking logs we see that $$\frac{6}{x} < \log(e^{6/x} + 8x) < \log 2 + \frac{6}{x}$$ Multiplying by $x/2$ we get $$3 < \frac{x\log(e^{6/x} + 8x)}{2} < \frac{x\log 2}{2} + 3$$ Letting $x \to 0^{+}$ we see that $(x/2)\log(e^{6/x} + 8x) \to 3$ and hence the given expression $(e^{6/x} + 8x)^{x/2} \to e^{3}$ as $x \to 0^{+}$.
A: Still take log: $\log f(x) = \dfrac{x\log(e^{6/x}+8x)}{2}$, and introduce $y = \dfrac{6}{x} \to f = f(y)$, and you can use L'hospitale rule.
