Determine constant a so that lim f(x) when x->0 exists when
f(x)={ arctan(1/x), x>0, (e^(ax)-1)/x, x<0 }
The answer should be pi/2, but i don't understand the reasoning behind it.
Anyone cares to explain?
Thanks
Let's rewrite so it is easier to read:
$\displaystyle f(x) = \begin{cases}\arctan\left( \dfrac{1}{x} \right) & x>0 \\ \dfrac{e^{ax}-1}{x} & x<0\end{cases}$
Now, take the limit from the left and the right as $x \to 0$. You want them to be equal in order for the limit to exist. So:
$\displaystyle \lim_{x \to 0^+}\arctan\left( \dfrac{1}{x} \right) = \dfrac{\pi}{2}$
So, you want:
$\displaystyle \dfrac{\pi}{2} = \lim_{x \to 0^-}\dfrac{e^{ax}-1}{x} = \lim_{x \to 0^-} ae^{ax} = a$
It is actually extremely straightforward.
L'Hospital's Rule.
Calculus I - L'Hospital's Rule and Indeterminate Forms