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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?

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@JanEerland gave the solution, but an approach without L'Hospital might be easier.

$$\frac{2x-3}{2x+5}=1-\frac8{2x+5}.$$

Then with $u=2x+5$,

$$\lim_{x\to\infty}\space\left(\frac{2x-3}{2x+5}\right)^{2x+1}=\lim_{u\to\infty}\left(1-\frac 8u\right)^{u-4}=\lim_{u\to\infty}\left(1-\frac 8u\right)^u\lim_{u\to\infty}\left(1-\frac 8u\right)^{-4}\\=e^{-8}\cdot1.$$

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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:

$$\text{L}=\lim_{x\to\infty}\space\left(\frac{2x-3}{2x+5}\right)^{2x+1}=\exp\left(-8\cdot1\right)=\frac{1}{e^8}$$

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  • $\begingroup$ What is the meaning of exp? $\endgroup$ – user6332995 Mar 5 '17 at 10:15
  • $\begingroup$ @user6332995 $$\exp\left(x\right)=e^x$$ $\endgroup$ – Jan Mar 5 '17 at 10:16
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    $\begingroup$ There is no real need for a second application of L'Hospital. Ratios of polynomials are trivial to solve (like you did next). $\endgroup$ – Yves Daoust Mar 5 '17 at 10:34
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    $\begingroup$ @JanEerland: why didn't you use it a third time, then ? $\endgroup$ – Yves Daoust Mar 5 '17 at 10:36
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    $\begingroup$ As you know, my point is that it was already obvious one step earlier. $\endgroup$ – Yves Daoust Mar 5 '17 at 10:39

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