# How to find this limit $\lim_{x \to \infty} \left(\frac{2x-3}{2x+5}\right)^{2x+1}$ using L'Hospital's Rule

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?

• $\log (a^b) = b \log (a)$ – Oria Gruber Mar 5 '17 at 10:06
• Yes 2x + 1 is the exponent of (2x-3/2x+5) – user6332995 Mar 5 '17 at 10:08
• – lab bhattacharjee Mar 5 '17 at 10:33
• – Workaholic Mar 5 '17 at 10:59

@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:

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

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