# Having trouble solving a limit

$$\lim _{x\to \infty }\left(\frac{2^{\frac{1}{x}}}{\left(2^{\frac{1}{x}}-1\right)x}\right)$$

I've tried a couple of things but I can't get to the right result.. according to online calculators the limit equals $$\frac{1}{ln\left(2\right)}$$ but I keep getting 0. Is there a a trick to solve it?

• Take the limits for the numerator and the denominator separately. Apr 12 at 11:32
• expand numerator and denumerator with $2^x$ Apr 12 at 11:38
• Thanks for the help guys! :) Apr 12 at 11:45

$$\lim _{x\to \infty }\left(\frac{2^{\frac{1}{x}}}{\left(2^{\frac{1}{x}}-1\right)x}\right)$$
taking $$1/x$$=$$t$$ as $$x$$ tends to $$\infty$$, $$t$$ tends to $$0$$
then apply L-hopital rule or use $$lim_{t\to0}\frac{2^t-1}{t}$$ = $$ln2$$