How to find $$\lim_{x \to \infty} \left(\frac{2x-3}{2x+5}\right)^{2x+1}$$
When I am calculating the limit I get a form like $\infty \times \infty$. I can't continue from that point. I made it as $\frac{\infty}0$. But L'Hospital's Rule can't apply here. Please help me to find the answer. Can you show me the way of doing that one?
@JanEerland gave the solution, but an approach without L'Hospital might be easier.
$$\frac{2x-3}{2x+5}=1-\frac8{2x+5}.$$
Then with $u=2x+5$,
$$\lim_{x\to\infty}\space\left(\frac{2x-3}{2x+5}\right)^{2x+1}=\lim_{u\to\infty}\left(1-\frac 8u\right)^{u-4}=\lim_{u\to\infty}\left(1-\frac 8u\right)^u\lim_{u\to\infty}\left(1-\frac 8u\right)^{-4}\\=e^{-8}\cdot1.$$
Well, we have that:
$$\text{L}=\lim_{x\to\infty}\space\left(\frac{2x-3}{2x+5}\right)^{2x+1}=\exp\left\{\lim_{x\to\infty}\space\frac{\frac{\text{d}}{\text{d}x}\left(\ln\left(\frac{2x-3}{2x+5}\right)\right)}{\frac{\text{d}}{\text{d}x}\left(\frac{1}{2x+1}\right)}\right\}=\exp\left\{\lim_{x\to\infty}\space\frac{\frac{16}{4x^2+4x-15}}{-\frac{2}{\left(1+2x\right)^2}}\right\}$$
And now you can use:
$$\frac{\frac{16}{4x^2+4x-15}}{-\frac{2}{\left(1+2x\right)^2}}=-8\cdot\frac{\left(1+2x\right)^2}{4x^2+4x-15}$$
So:
$$\text{L}=\exp\left\{-8\lim_{x\to\infty}\space\frac{\frac{\text{d}}{\text{d}x}\left(\left(1+2x\right)^2\right)}{\frac{\text{d}}{\text{d}x}\left(4x^2+4x-15\right)}\right\}=\exp\left\{-8\lim_{x\to\infty}\space\frac{4\left(1+2x\right)}{4+8x}\right\}$$
And now you can use:
$$\frac{4\left(1+2x\right)}{4+8x}=1$$
So:
$$\text{L}=\lim_{x\to\infty}\space\left(\frac{2x-3}{2x+5}\right)^{2x+1}=\exp\left(-8\cdot1\right)=\frac{1}{e^8}$$
