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$\frac{log (1+\frac{x}{p})- log(1-\frac{x}{q})}{x}$ $,x\neq 0$ The function is continuous at $x=0$ , and we need to find $f(0)$.

The numerator goes to $0$ $(log1 - log 1)$ And the denominator goes to $0$ as well. So $f(0)= 0/0=1$ Is this correct?

2 Answers 2

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Hint: compute $\lim_{x \to 0}\frac{log (1+\frac{x}{p})- log(1-\frac{x}{q})}{x}$ with l'Hospital

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    Oh! I got it, the answer is $\frac{1}{p}+\frac{1}{q}$? – 2017-01-12
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    Your answer is correct. – 2017-01-12
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You can also use equivalents.

Since, for small values of $\epsilon$, $\log(1+\epsilon)\sim \epsilon$, then $$\log (1+\frac{x}{p})\sim \frac{x}{p}$$ $$\log (1-\frac{x}{q})\sim -\frac{x}{q}$$ $$\log (1+\frac{x}{p})-\log (1-\frac{x}{q})\sim \frac{x}{p}+\frac{x}{q}=x\left(\frac{1}{p}+\frac{1}{q} \right)$$