I need to calculate the following limit without using L'Hôpital's rule:
$$ \lim_{x\to\infty} \big(\frac{x}{2}\big)^{\frac{1}{x-2}} $$
I written the expression using $ln$:
$$ \lim_{x\to\infty} e^{\frac{ln(\frac{x}{2})}{x-2}} $$
I don't know how calculate the limit of the exponent (without L'Hôpital and without derivative at all). How can I do it?
Hint. Make the substitution $$ y = \frac{1}{x-2}, $$ solve for $x$ and think about the definition of $e$ as a limit.
\begin{eqnarray} \lim_{x\to\infty} \left(\frac{x}{2}\right)^{\frac{1}{x-2}}&=&\lim_{x\to\infty}\exp\left(\frac{\ln(x)-\ln(2)}{x-2}\right)\\ &=&\exp\left(\lim_{x\to\infty}\frac{\ln(x)-\ln(2)}{x-2}\right)\\ &=&e^0\\ &=&1 \end{eqnarray}
You can simply squeeze it by using the generalized Bernoulli inequality for exponents between $0$ and $1$. So for $\bf{x>3}$ you have
$$1\leq \left(\frac x2\right)^{\frac 1{x-2}}\leq x^{\frac 1{x-2}}\leq (1+x)^{\frac 1{x-2}}\stackrel{Bernoulli}{\leq}1+\frac{x}{x-2}\stackrel{x\to\infty}{\longrightarrow}1$$
