$ 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!
$ 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!
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.