Approximation of a hyperbolic function

[Taylor_1989]

Homework Statement

Hy guys I am having an issue with approximating this first question, which I have shown below.

Now my problem is not so much solving it but I have been thinking that if given the same question without knowing that it approximates to so for example the question I am thinking of is: approximate the function: $\frac{cosh x}{sinh x}-1$.

Homework Equations

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

The Attempt at a Solution

[/B]
So alls I have at the moment is $\lim_{x\to\infty}\frac{2e^{-2x}}{1-2e^{-2x}}$
and this is where I get stuck how would I use limits to show the it approximates to what they have given. I was thinking of just taking the limit of the bottom line, and this would give the ans but I am not 100% sure that you can actually do that, another way I was thinking is showing what the end of the function dose, but then I would have to dived through and I am not sure that the function can be divided. Could someone please give me some advice much appreciated.

 

[Buffu]

I think you probably use @Taylor_1989 series and inequalities.

 

[SammyS]

Homework Statement

Hy guys I am having an issue with approximating this first question, which I have shown below.

View attachment 198780

Now my problem is not so much solving it but I have been thinking that if given the same question without knowing that it approximates to so for example the question I am thinking of is: approximate the function: $\frac{cosh x}{sinh x}-1$.

Homework Equations

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

The Attempt at a Solution

[/B]
So alls I have at the moment is $\lim_{x\to\infty}\frac{2e^{-2x}}{1-2e^{-2x}}$
and this is where I get stuck how would I use limits to show the it approximates to what they have given. I was thinking of just taking the limit of the bottom line, and this would give the ans but I am not 100% sure that you can actually do that, another way I was thinking is showing what the end of the function dose, but then I would have to dived through and I am not sure that the function can be divided. Could someone please give me some advice much appreciated.

There's an error in the denominator of

$\displaystyle\frac{2e^{-2x}}{1-2e^{-2x}}$

 

[scottdave]

So we have 2*cosh(x) = [e^x + e^-x] and 2*sinh(x) = [e^x - e^-x]. You would agree that cosh(x) / sinh(x) is equal to (2*cosh) / (2*sinh). Now replace the 1 with (2*sinh) / (2*sinh), so you now have : [e^x + e^-x - (e^x - e^-x)] / [e^x - e^-x]. With a little rearranging, you may be able to simplify it.

 

[scottdave]

So we have 2*cosh(x) = [e^x + e^-x] and 2*sinh(x) = [e^x - e^-x]. You would agree that cosh(x) / sinh(x) is equal to (2*cosh) / (2*sinh). Now replace the 1 with (2*sinh) / (2*sinh), so you now have : [e^x + e^-x - (e^x - e^-x)] / [e^x - e^-x]. With a little rearranging, you may be able to simplify it.

Or actually with what you have, the denominator of 1 - 2*e^(-2x) approaches 1 as x gets large. So you can say it is equal to 2*e^(-2x) in the numerator, and approximately 1 in the denominator, so it approaches 2*e^(-2x)