If the function f(x)=cases1x_e(1+x/a1-x/b), & x0cases is cont…

Question

If the function $f(x)=\begin{cases}\dfrac1x\log_e\left(\dfrac{1+x/a}{1-x/b}ight), & x<0\k, & x=0\\dfrac{\cos2x-\sin2x-1}{\sqrt{x^2+1}-1}, & x>0\end{cases}$ is continuous at $x=0$, then $\dfrac1a+\dfrac1b+\dfrac4k$ is equal to

Accepted Answer

LHL: $\displaystyle\lim_{x	o0^-}\frac1x\ln\left(1+\frac xaight)-\frac1x\ln\left(1-\frac xbight)=\frac1a+\frac1b$.

RHL: $\displaystyle\lim_{x	o0^+}\frac{-2\sin2x}{\sqrt{x^2+1}-1}=\lim_{x	o0^+}\frac{-2\sin2x}{x^2}\left(\sqrt{x^2+1}+1ight)=-4$.

So $k=-4=\dfrac1a+\dfrac1b$, giving $\dfrac1a+\dfrac1b+\dfrac4k=-4+\dfrac{4}{-4}=-4-1=-5$.

Solutions · 1