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$$ f(x) = \begin{cases} \frac{1-\cos(5x)}{x^2} & x > 0 \\ \\ \frac{e^x + 2x -2}{x} & x < 0 \end{cases} $$

Find the limits for $x\rightarrow 0^-$ and for $x\rightarrow 0^+$.

Thanks in advance for any help!

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Do you know what it means $x\rightarrow 0^-$ and $x\rightarrow 0^{+}$? –  Mhenni Benghorbal Dec 22 '12 at 20:38
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1 Answer

up vote 1 down vote accepted

Hints: $$\lim_{x\to 0^+}\frac{1-\cos 5x}{x^2}=\lim_{u\to 0^+}\frac{1-\cos u}{(\frac u 5)^2}$$ while $$\lim_{x\to 0^-}\frac{e^x+2x-2}{x}$$ is not indeterminate.

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