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I tried:

$$\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x} = \\ \frac{e^{x}(1+e^{-2x}-\frac{2}{e^x})}{x(x-2)} = \frac{e^x(1+e^{-2x})-2}{x(x-2)} = \frac{e^x(1+e^{-2x})}{x(x-2)} - \frac{2}{x(x-2)} = \\ \frac{1+e^{-2x}}{x} \cdot \frac{e^x}{x-2} - \frac{2}{x(x-2)} = ???$$

What do I do next?

  • 0
    maybe this can help: recall that $cosh(x) = \frac{e^x+e^{-x}}{2}$ – 2017-02-21
  • 1
    The factorization of the denominator is wrong should be $x(x + 2)$ – 2017-02-21

4 Answers 4

1

Simply define $f(x) = e^x + e^{-x}.$ Then the expression equals

$$\frac{f(x) - f(0)}{x-0}\cdot \frac{x}{x^2+2x}.$$

As $x\to0,$ the first fraction $\to f'(0)$ by the definition of the derivative. The second fraction $\to 1/2.$ Since $f'(0)=0,$ the limit in question is $0\cdot (1/2) = 0.$

  • 0
    This presupposes that we know that $e^x$ is differentiable though. – 2017-02-21
  • 0
    That's a strange comment. I found the limit without L'Hopital, which is what the OP indicates is fair play. How else would you have us proceed? – 2017-02-21
  • 0
    Oh I would have proceeded in the same way. I just know that at my university students are sometimes asked such questions before differentiability is dealt with. So it was just a remark for future students, not a critique of your answer. I agree that this method is perfectly in line with what the OP has specified. – 2017-02-21
7

You can use hyperbolics: $2\cosh x=e^x+e^{-x}$ and so the limit expression can be written as $$\frac{2}{x+2}\cdot\frac{\cosh x-1}{x-0}$$ and use the limit definition of the derivative on $y=\cosh x$ taken at $x=0$. Now give it a try from here.

  • 0
    To format $2coshx$ correctly, you need to write `2\cosh x` = $2\cosh x$ – 2017-02-21
  • 2
    Better yet, since $\cosh(x)$ is even and defined at $x=0$, it's derivative is zero.$$\cosh'(x)=-\cosh'(-x)\implies\cosh'(0)=0$$ – 2017-02-21
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    @amWhy Thanks, will take note of it...My formatting is still a learning experience. – 2017-02-22
5

The path you took leads nowhere, because you get a $\infty-\infty$ form. (By the way, $x^2+2x=x(x+2)$, not $x(x-2)$, but it's a detail.)

Use the fact that $$ e^x+e^{-x}-2=e^{-x}(e^x-1)^2 $$ so your limit can be written as $$ \lim_{x\to0}\frac{e^{-x}}{x+2}\frac{e^x-1}{x}(e^x-1) $$

Of course, l’Hôpital in this case is much easier:

$$\displaystyle\lim_{x\to0}\frac{e^x-e^{-x}}{2x+2}$$

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    Ugh! You beat me to it! – 2017-02-21
  • 0
    @SimplyBeautifulArt On the other hand, this is a case where l'Hôpital is *much* easier! – 2017-02-21
  • 0
    I could solve it this way $\frac{e^x+e^{-x}-2}{x^2+2x} = \frac{e^{-x}(e^x-1)^2}{x(x+2)} = \frac{(e^x-1)^2}{x} \cdot \frac{e^{-x}}{x+2} = (e^x-1) \cdot \frac{e^x-1}{x} \cdot \frac{e^{-x}}{x+2} = 0 \cdot 1 \cdot \frac{1}{2} = 0$. Thanks. I have one more question though. How did you know that $e^x+e^{-x}-2=e^{-x}(e^x-1)^2$? – 2017-02-21
  • 0
    @MarkRead Experience? `;-)` It's $t-2+1/t=(t^2-2t+1)/t$ – 2017-02-21
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    Ok, what "rules"/properties did you use then? – 2017-02-21
  • 0
    @MarkRead See above – 2017-02-21
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    Is $t$ equal to $e^x$ or $e^{-x}$? – 2017-02-21
  • 0
    @MarkRead $t=e^x$. – 2017-02-21
  • 0
    Right, I understand now. Another way of representing this formula would be $t-2+t^{-1} = (t^2-2t+1)\cdot t^{-1}$. Can 2 be replaced by any constant? Or does it have to be specifically 2? – 2017-02-21
3

Aside from not carelessly turning $x(x+2)$ into $x(x-2)$, partial fractions is the wrong tool for this job. If you are allowed to used Taylor series, you can see that $$ e^x + e^{-x} - 2 = x^2 + O(x^x)$$ so you get the expression $$\frac{x^2}{x(x+2)} = \frac{x}{x+2} \to 0 $$