1. ## Limit problem

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

2. ## Re: Limit problem

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.

3. ## Re: Limit problem

Originally Posted by SlipEternal

It is actually extremely straightforward.
Thank you! That's extremely helpful, also thanks for formatting, i'll steal it in the future

4. ## Re: Limit problem

Sorry, my math is quite rusty, how did you lower the exponent a?

$\displaystyle \dfrac{\pi}{2} = \lim_{x \to 0^-}\dfrac{e^{ax}-1}{x} = \lim_{x \to 0^-} ae^{ax} = a$

6. ## Re: Limit problem

Or recognise the limit definition of the derivative at zero.