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Find $a,c$ such that:

$$f(x)= \begin{cases} a\frac{\exp(tgx)}{(1+\exp(tgx))} &\text{for }|x|<\pi/2 \\[2ex] \exp(cx)-2 &\text{for } |x|\ge\pi/2 \end{cases}$$

is continuous.

How do I evaluate the left- and right-hand limits to see if they are equal?

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exp(tgx) is equal to $e^(tanx)$? – dwarandae Jan 16 '13 at 20:26

Note that $$ \lim_{x\to(\frac\pi2)^-}\frac{a\exp\tan x}{1+\exp\tan x}=\lim_{x\to+\infty}\frac{a\exp x}{1+\exp x}=\lim_{x\to+\infty}\frac{a x}{1+ x}=a.$$

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ok then $exp(c*(\frac{\pi}{2}))-2=a \implies c=\frac{2ln(2+a)}{\pi}$ But have either $ lim_{x\to-\pi/2}$ – aiki93 Jan 16 '13 at 21:11

To prove the function is continuous, you need to fullfill the condition

$$ \lim_{x\to x_0}f(x)=f(x_0). $$

So, as a first step, you need to prove that the limit exists at the point $x_0$, that is why you need to prove $$\lim_{x\to \pi/2^{-}}f(x)= \lim_{x\to \pi/2^{+}}f(x).$$

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