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I am a Calc 2 student and am having trouble seeing this limit, the assignment is to use the definition of the hyperbolic functions to find the limit.

$\lim_{x\to 0+} Coth(x)$

the answer is $+\infty$

If I work it out I get to this point: $lim_{x\ to 0+}$ $(e^(2x)+1)/(e^(2x) -1)$

Please explain it using the definition of the Hyperbolic functions.

Work:

$lim_{x\ to 0+}$ $cosh(x)/sinh(x)$ = $lim_{x\ to 0+}$ $e^(x)+e^(-x))/e^(x)-e^(-x)$

$lim_{x\ to 0+}$ $e^x(e^(x)+e^(-x))/e^(x)-e^(-x))$

$lim_{x\ to 0+}$ $e^(2x)+1)/e^(2x)-1$ = (1+1)/(1-1) = 2/0

I don't see how 2/0 relates to 1/x like everyone keeps saying, I am extremely frustrated at this point as my professor couldn't explain it without using the graph. If i knew what the graph looked like this would be pointless. Please I am missing some fundamental idea of limits or something.

Ryan
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3 Answers3

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First remark : $\lim_{x \to 0^+}\cosh (x)=1$ and not $2$ as you wrote (typo, I guess).

You properly wrote $$\coth(x)=\frac{e^{2x}+1}{e^{2x}-1}$$ Now, remember that, for small $y$, $$e^y=1+y+O\left(y^2\right)$$ so $$e^{2x}=1+2 x+O\left(x^2\right)$$ and then, for small $x$ $$\coth(x)=\frac{1+2 x+O\left(x^2\right)+1}{1+2 x+O\left(x^3\right)-1}=\frac{2+2 x+O\left(x^2\right)}{2 x+O\left(x^2\right)}\sim \frac 1x$$

Claude Leibovici
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  • Thank you for responding, although Google search brings up nothing regarding e^y = 1 + y + 0(y^2). I have since edited the question to show my work please look it over. – Ryan Feb 04 '16 at 06:30
  • @Ryan. Have a look at https://en.wikipedia.org/wiki/Exponential_function – Claude Leibovici Feb 04 '16 at 06:33
  • Thankyou, that did put me on the right path to understanding what was happening. I just didnt remember limits correctly. – Ryan Feb 09 '16 at 03:52
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Observe that, for any $a>0$, $$ \lim_{x \to 0^+}\frac{a}{x}=+\infty $$ giving $$ \lim_{x \to 0^+}\coth x=\lim_{x \to 0^+}\frac{1}{x}=+\infty. $$

Olivier Oloa
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  • When I work this out using the def of hyperbolics I get a finite # in the numerator and the denominator. When I plug in 0 into e^2(x) I get 1, 1-1 = 0, if it was purely just x I could see your example. Am I to ignore the e^2 completely as it is infinitesimal compared to infinity? Am I to attach the superscript of + to the resultant 1 from e^(2x)? I hope I was able to convey that correctly. IN your first example replace the x of the denominator with a f(x), does that definition still hold? please I need the logic behind this sorted out, please explain this like I am 5 – Ryan Feb 03 '16 at 04:30
  • Please, you may consider here the Taylor series, giving $$\coth x = \frac{\cosh x}{\sinh x} \sim \frac1{x}$$ as $x \to 0^+$. – Olivier Oloa Feb 03 '16 at 04:35
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So from this point -> $\lim_{x \to 0+} (e^{2x}+1)/(e^{2x}−1)$

If you consider the top and bottom bits separately, the top one is always going to be almost 2, whereas the bottom one will be an infinitely small number greater than $0 - e^{2x}$ for an infinitely small $x$ is just over $1$ (as $e^{2x}$ for $x=0$ is $1$). Hence you get $2/0+$, which is +∞.

Note: if $x$ goes to $0-$, you get $e^{2x}$ being just under $1$, so you would have the limit being $2/0-$, which is $-∞$.

Hanul Jeon
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Spas Banchev
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