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How can I find this limit

$$ \lim_{x \to 0}\frac{e^{2x}-3e^{x}+2}{5x} $$

without using L'Hopital's rule?


I know this is true: $$ \lim_{x \to 0}\frac{e^{x}-1}{x} = 1 $$ So I've tried to isolate the $5x$ so that $(1/5)$ multiplies by all of it. Then I'm not sure what to do.

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    $\begingroup$ What have you tried so far? Try what was suggested in the question you asked $30$ minutes ago math.stackexchange.com/questions/1139231/… and recreate the expression $\frac{e^x-1}{x}$ from what was given. $\endgroup$ – graydad Feb 8 '15 at 16:28
  • $\begingroup$ I've tried to isolate the 5x so that (1/5) multiplies by all of it. Then i'm not sure what to do .. $\endgroup$ – João Silva Feb 8 '15 at 16:30
  • $\begingroup$ Maybe the numerator can be factored? $\endgroup$ – graydad Feb 8 '15 at 16:31
  • $\begingroup$ If you're familiar with Taylor/Mclaurin expansions you can expand the exponential terms to first order near 0. $\endgroup$ – Jack Feb 8 '15 at 16:34
  • $\begingroup$ What about $e^{2x}-3e^x+2=(e^{2x}-1)-3(e^x-1)$? $\endgroup$ – Martin Sleziak Feb 8 '15 at 17:06
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HINT

$$e^{2x}-3e^x+2=(e^x)^2-3e^x+2=(e^x-1)(e^x-2)$$

$$\implies\frac{e^{2x}-3e^x+2}{5x}=\frac15\cdot\frac{e^x-1}x(e^x-2)$$

Can you take it home from here?

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  • $\begingroup$ Yes, thanks i managed to solve it with that! $\endgroup$ – João Silva Feb 8 '15 at 16:36
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Hint: $$\displaystyle\frac{e^{2x} - 3 e^x + 2}{5x}=\frac 15\left(e^x\frac{e^x-1}{x} - 2\frac{e^x-1}x\right)$$

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$$\lim_{x\to 0}\frac{e^{2x}-3e^x+2}{5x}$$ $$=\frac{1}{5}\lim_{x\to 0}\frac{\left(1+\frac{(2x)}{1!}+\frac{(2x)^2}{2!}+\frac{(2x)^3}{3!}+.... \infty\right)-3\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.... \infty\right)+2}{x}$$ $$=\frac{1}{5}\lim_{x\to 0}\frac{2x\left(\frac{(2x)}{2!}+\frac{(2x)^2}{3!}+.... \infty\right)-3x\left(\frac{x}{2!}+\frac{x^2}{3!}+.... \infty\right)+(2x-3x)}{x}$$ $$=\frac{1}{5}\lim_{x\to 0} \left[2 \left(\frac{(2x)}{2!}+\frac{(2x)^2}{3!}+.... \infty\right)-3\left(\frac{x}{2!}+\frac{x^2}{3!}+.... \infty\right)-1\right]$$ $$=\frac{1}{5}\left[-1\right]=-\frac{1}{5}$$

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