Can you assume e^-x approaches 0 when finding the limit of tanh(x)?

[pierce15]

Here it is:
$$
\lim_{x\to\infty} tanh(x)=\lim_{x\to\infty} sinh(x)/cosh(x)=\lim_{x\to\infty} \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}
$$

Can you just assume that the $$e^{-x}$$ will approach 0 and the $$e^{x}$$ will cancel and the limit will reduce to 1?

 

[micromass]

Well, it's correct. But perhaps your teacher might want a bit more formal and rigorous solution? Depends on your teacher of course.

 

[pierce15]

Well, it's correct. But perhaps your teacher might want a bit more formal and rigorous solution? Depends on your teacher of course.

That's what I was wondering, what might be a more rigorous approach to this?

 

[pierce15]

By the way, I'm not sure why this was moved because it's not homework

Nonetheless, I found a way to do this:

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

The second limit is simplified enough that we know that it is equivalent to 0.

$$lim_{x\to\infty} \frac{e^{x}}{e^{x}+e^{-x}}$$
$$lim_{x\to\infty} \frac{1}{\frac{e^{x}+e^{-x}}{e^{x}}}$$
$$lim_{x\to\infty} \frac{1}{1+e^{-2x}}$$

This is enough to show that the first limit is equivalent to 1.

 

[Mark44]

By the way, I'm not sure why this was moved because it's not homework

Nonetheless, I found a way to do this:

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

It's much simpler just to factor e^x out of the expressions in the numerator and denominator.

$$ \lim_{x\to\infty} \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$
$$ = \lim_{x\to\infty} \frac{e^x(1 - e^{-2x})}{e^x(1 + e^{-2x})} $$

Also, you should connect your limit expressions with =.

$$lim_{x\to\infty} \frac{e^{x}}{e^{x}+e^{-x}}-lim_{x\to\infty} \frac{e^{x}}{e^{x}+e^{-x}}-\frac{e^{-x}}{e^{x}+e^{-x}}$$\frac{e^{-x}}{e^{x}+e^{-x}}[/tex]
$$lim_{x\to\infty} \frac{e^{x}}{e^{x}+e^{-x}}-lim_{x\to\infty} \frac{e^{x}}{e^{x}+e^{-x}}-\frac{e^{-x}}{e^{x}+e^{-x}}$$\frac{1}{e^{x}(e^{x}+e^{-x})}[/tex]
$$lim_{x\to\infty} \frac{e^{x}}{e^{x}+e^{-x}}-lim_{x\to\infty} \frac{e^{x}}{e^{x}+e^{-x}}-\frac{e^{-x}}{e^{x}+e^{-x}}$$\frac{1}{e^2x}+1)}[/tex]

The second limis simplified enough that we know that it is equivalent to 0.

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

Fixed the LaTeX in the one below.

$$lim_{x\to\infty} \frac{1}{\frac{e^x+e^{-x}}{e^x}}$$
$$lim_{x\to\infty} \frac{1}{1+e^{-2x}}$$

This is enough to show that the first limit is equivalent to 1.

 

[Michael Redei]

How about trying de L'Hopital's Rule, after you've multiplied the numerator and denominator by e^x?
$$\lim\frac{ e^x-e^{-x} }{ e^x+e^{-x} } =<br /> \lim\frac{ e^{2x}-1 }{ e^{2x}+1 } =<br /> \lim\frac{ 2e^{2x} }{ 2e^{2x} } =<br /> \lim1 = 1.$$

 

[micromass]

By the way, I'm not sure why this was moved because it's not homework

That doesn't matter. It's a homework-style question from a textbook, so it belongs in the homework forums. This is regardless of that it actually is homework or is self-study.

 

[pierce15]

How about trying de L'Hopital's Rule, after you've multiplied the numerator and denominator by e^x?
$$\lim\frac{ e^x-e^{-x} }{ e^x+e^{-x} } =<br /> \lim\frac{ e^{2x}-1 }{ e^{2x}+1 } =<br /> \lim\frac{ 2e^{2x} }{ 2e^{2x} } =<br /> \lim1 = 1.$$

$$That\space would\space have\space saved\space me\space a\space lot\space of\space time...$$

 

[Mark44]

$$That\space would\space have\space saved\space me\space a\space lot\space of\space time...$$

As would have factoring e^x out of the top and bottom.

Also, why did you write the text above as LaTeX? LaTeX is very useful for depicting fractions and integrals and such, but some posts with large amounts of LaTeX take a long time to render in some browsers.

 

[pierce15]

As would have factoring e^x out of the top and bottom.

Also, why did you write the text above as LaTeX? LaTeX is very useful for depicting fractions and integrals and such, but some posts with large amounts of LaTeX take a long time to render in some browsers.

Because Michael was helping me with latex before