Finding hypberbolic/exponential limit

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In summary, the problem was to find the limit of (cosh(2x))/(e^(2x)). The solution involves using the identity cosh(x)=(e^x+e^-x)/2 and then taking the limit as x approaches infinity. After some algebraic manipulations, the limit is simplified to 1/2.
  • #1
silicon_hobo
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[SOLVED] finding hypberbolic/exponential limit

Homework Statement


Find:
[tex]\mathop{\lim} \limits_{x \to \infty} \frac{cosh(2x)}{e^2^x}[/tex]

Homework Equations


[tex]cosh = \frac{e^x+e^-^x}{2}[/tex]

The Attempt at a Solution


[tex]\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[/tex]

Hey, it's me again. This latex script is groovy. Am I on the right track here? Cheers.
 
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  • #2
[tex]\lim_{x\rightarrow\infty}\frac{\cosh{2x}}{e^{2x}}[/tex]

[tex]\lim_{x\rightarrow\infty}\frac{e^{2x}+e^{-2x}}{2e^{2x}}[/tex]

[tex]\frac 1 2\lim_{x\rightarrow\infty}\left(\frac{e^{2x}}{e^{2x}}+\frac{e^{-2x}}{e^{2x}}\right)[/tex]

Take it from here ...
 
Last edited:
  • #3
I have no idea what
[tex]\frac{\ln}{\ln} (\frac{e^2^x+e^-^2^x}{2e^2^x}) \[/tex]
means!
 
  • #4
Thanks for the replies.

HallsofIvy said:
I have no idea what
[tex]\frac{\ln}{\ln} (\frac{e^2^x+e^-^2^x}{2e^2^x}) \[/tex]
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.

rocophysics said:
[tex]\lim_{x\rightarrow\infty}\frac{\cosh{2x}}{e^{2x}}[/tex]

[tex]\lim_{x\rightarrow\infty}\frac{e^x+e^{-x}}{2e^{2x}}[/tex]

[tex]\frac 1 2\lim_{x\rightarrow\infty}\left(\frac{e^x}{e^{2x}}+\frac{e^{-x}}{e^{2x}}\right)[/tex]

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:

[tex]\lim_{x\rightarrow\infty}\frac{e^{2x}+e^{-2x}}{2e^{2x}}[/tex]

[tex]\frac 1 2\lim_{x\rightarrow\infty}\left(\frac{e^2^x}{e^{2x}}+\frac{e^{-2x}}{e^{2x}}\right)[/tex]

and then maybe...

[tex]\frac 1 2\lim_{x\rightarrow\infty}\left(2\right) = 1[/tex]

?
 
  • #5
SHOOT! So sorry :) Thank you for fixing it ... I was so caught up on the LaTeX, lol.

So your answer is 1/2
 
  • #6
rocophysics said:
So your answer is 1/2

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

Thanks!
 
  • #7
I pulled out the 1/2 infront of the limit.

[tex]\frac 1 2\lim_{x\rightarrow\infty}\left(\frac{e^{2x}}{e^{2 x}}+\frac{e^{-2x}}{e^{2x}}\right)[/tex]

[tex]\frac 1 2\lim_{x\rightarrow\infty}\left(1+\frac{1}{e^{4x}}\right)[/tex]

[tex]\frac 1 2\lim_{x\rightarrow\infty}\left(1+0\right)[/tex]
 
  • #8
Aha! I've got it now. Thanks.
 

1. What is a hyperbolic limit?

A hyperbolic limit is a mathematical concept that describes the behavior of a function at its critical points. It is used to determine the values that a function approaches as its input approaches a specific value, often represented as x → a.

2. How is a hyperbolic limit different from an exponential limit?

A hyperbolic limit is based on the behavior of a hyperbolic function, such as a hyperbolic sine or cosine, while an exponential limit is based on the behavior of an exponential function, such as e^x. The key difference is that hyperbolic functions have a horizontal asymptote at y = 0, while exponential functions have a horizontal asymptote at y = 1.

3. What are some common methods for finding hyperbolic/exponential limits?

Some common methods for finding hyperbolic/exponential limits include using L'Hopital's rule, graphing the function, and using algebraic manipulation to simplify the expression. Additionally, understanding the properties and behavior of hyperbolic and exponential functions can also be helpful in finding these limits.

4. Can hyperbolic/exponential limits have different values from left and right limits?

Yes, hyperbolic/exponential limits can have different values from left and right limits. This can occur if the function approaches different values from the left and right sides of the critical point. In these cases, the overall limit does not exist.

5. In what real-life scenarios are hyperbolic/exponential limits useful?

Hyperbolic/exponential limits are often used in physics and engineering to model real-life situations, such as the motion of a spring or the growth of a population. They can also be used in financial analysis to predict the behavior of investments over time. Essentially, any situation that involves exponential or hyperbolic growth or decay can benefit from understanding and finding these limits.

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