# Find the limit $\lim_{x \to 0}\frac{e^{2x}-3e^{x}+2}{5x}$ without using L'Hopital rule.

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.

• 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. – graydad Feb 8 '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 8 '15 at 16:30
• Maybe the numerator can be factored? – graydad Feb 8 '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 8 '15 at 16:34
• What about $e^{2x}-3e^x+2=(e^{2x}-1)-3(e^x-1)$? – Martin Sleziak Feb 8 '15 at 17:06

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?

• Yes, thanks i managed to solve it with that! – João Silva Feb 8 '15 at 16:36

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

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