Calculate the limit: $\lim_{x\to \infty} (1-\frac{2x+1}{2x-1})^x$

Question

how can I solve this limit without using l'hopital.

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

Answer 1

$$\left(1-\frac{2x+1}{2x-1}\right)^x=\left(-\frac2{2x-1}\right)^x=\ldots$$

the expression isn't well defined for infinite values of $\;x\;$ when $\;x \to\infty\;$ and thus the limit cannot be taken.

Answer 2

$$ \left(1-\frac{2x+1}{2x-1}\right)^x=\left(\frac{-2}{2x-1}\right)^x$$

thus the limit exist only for $x\to -\infty$, set $y=-x\to +\infty$

$$\left(\frac{-2}{2x-1}\right)^x=\left(\frac{2}{2y+1}\right)^{-y}=e^{-y\log{\frac{2}{2y+1}}}\to +\infty$$