Finding hypberbolic/exponential limit

[silicon_hobo]

[SOLVED] finding hypberbolic/exponential limit

Homework Statement

Find:
$$\mathop{\lim} \limits_{x \to \infty} \frac{cosh(2x)}{e^2^x}$$

Homework Equations

$$cosh = \frac{e^x+e^-^x}{2}$$

The Attempt at a Solution

$$\mathop{\lim} \limits_{x \to \infty} \frac{cosh(2x)}{e^2^x} \rightarrow \frac{\frac{e^2^x+e^-^2^x}{2}}{e^2^x} \rightarrow \frac{e^2^x+e^-^2^x}{2}*\frac{1}{e^2^x} \rightarrow \frac{\ln}{\ln} (\frac{e^2^x+e^-^2^x}{2e^2^x}) \rightarrow \frac{2x}{2x\ln2} - \frac{2x}{2x\ln2} = 0$$

Hey, it's me again. This latex script is groovy. Am I on the right track here? Cheers.

 

[rocomath]

$$\lim_{x\rightarrow\infty}\frac{\cosh{2x}}{e^{2x}}$$

$$\lim_{x\rightarrow\infty}\frac{e^{2x}+e^{-2x}}{2e^{2x}}$$

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

Take it from here ...

 

[HallsofIvy]

I have no idea what
$$\frac{\ln}{\ln} (\frac{e^2^x+e^-^2^x}{2e^2^x}) \$$
means!

 

[silicon_hobo]

Thanks for the replies.

I have no idea what
$$\frac{\ln}{\ln} (\frac{e^2^x+e^-^2^x}{2e^2^x}) \$$
means!

I thought I should take the ln of the fraction to eliminate the exponents. However, rocophysics' reply indicates this strategy was incorrect. Unfortunately, I am still confused.

$$\lim_{x\rightarrow\infty}\frac{\cosh{2x}}{e^{2x}}$$

$$\lim_{x\rightarrow\infty}\frac{e^x+e^{-x}}{2e^{2x}}$$

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

Take it from here ...

On account of the cosh(2x) in the original problem, I believe steps 2 and 3 quoted above should read as follows:

$$\lim_{x\rightarrow\infty}\frac{e^{2x}+e^{-2x}}{2e^{2x}}$$

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

and then maybe...

$$\frac 1 2\lim_{x\rightarrow\infty}\left(2\right) = 1$$

?

 

[rocomath]

SHOOT! So sorry :) Thank you for fixing it ... I was so caught up on the LaTeX, lol.

So your answer is 1/2

 

[silicon_hobo]

So your answer is 1/2

Hmmm. If $$\left(\frac{e^2^x}{e^{2x}}+\frac{e^{-2x}}{e^{2x}}\right)=2$$ shouldn't the answer be 1?

Thanks!

 

[rocomath]

I pulled out the 1/2 infront of the limit.

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

$$\frac 1 2\lim_{x\rightarrow\infty}\left(1+\frac{1}{e^{4x}}\right)$$

$$\frac 1 2\lim_{x\rightarrow\infty}\left(1+0\right)$$

 

[silicon_hobo]

Aha! I've got it now. Thanks.