- #1
silicon_hobo
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[SOLVED] finding hypberbolic/exponential limit
Find:
[tex]\mathop{\lim} \limits_{x \to \infty} \frac{cosh(2x)}{e^2^x}[/tex]
[tex]cosh = \frac{e^x+e^-^x}{2}[/tex]
[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.
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.