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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.

João Silva
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    What have you tried so far? Try what was suggested in the question you asked $30$ minutes ago http://math.stackexchange.com/questions/1139231/limit-solving-without-using-lhopital-rule

    and recreate the expression $\frac{e^x-1}{x}$ from what was given.

     – graydad Feb 08 '15 at 16:28
  • I've tried to isolate the 5x so that (1/5) multiplies by all of it. Then i'm not sure what to do .. – João Silva Feb 08 '15 at 16:30
  • Maybe the numerator can be factored? – graydad Feb 08 '15 at 16:31
  • If you're familiar with Taylor/Mclaurin expansions you can expand the exponential terms to first order near 0. – Jack Feb 08 '15 at 16:34
  • What about $e^{2x}-3e^x+2=(e^{2x}-1)-3(e^x-1)$? – Martin Sleziak Feb 08 '15 at 17:06

3 Answers3

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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?

lab bhattacharjee
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Hint: $$\displaystyle\frac{e^{2x} - 3 e^x + 2}{5x}=\frac 15\left(e^x\frac{e^x-1}{x} - 2\frac{e^x-1}x\right)$$

Mathmo123
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$$\lim_{x\to 0}\frac{e^{2x}-3e^x+2}{5x}$$ $$=\frac{1}{5}\lim_{x\to 0}\frac{\left(1+\frac{(2x)}{1!}+\frac{(2x)^2}{2!}+\frac{(2x)^3}{3!}+.... \infty\right)-3\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.... \infty\right)+2}{x}$$ $$=\frac{1}{5}\lim_{x\to 0}\frac{2x\left(\frac{(2x)}{2!}+\frac{(2x)^2}{3!}+.... \infty\right)-3x\left(\frac{x}{2!}+\frac{x^2}{3!}+.... \infty\right)+(2x-3x)}{x}$$ $$=\frac{1}{5}\lim_{x\to 0} \left[2 \left(\frac{(2x)}{2!}+\frac{(2x)^2}{3!}+.... \infty\right)-3\left(\frac{x}{2!}+\frac{x^2}{3!}+.... \infty\right)-1\right]$$ $$=\frac{1}{5}\left[-1\right]=-\frac{1}{5}$$

Harish Chandra Rajpoot
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