Given that
f(x)={\begin{matrix}\frac{1}{x}{\mathrm{log}}_{e}(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}),x<0 \\ k,x=0 \\ \frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x-1}{\sqrt{{x}^{2}+1}-1},x>0\end{matrix}
f ( x ) f(x) f ( x ) x = 0 x=0 x = 0
L . H . L = R . H . L = f ( 0 ) L.H.L=R.H.L=f(0) L . H . L = R . H . L = f ( 0 )
L H L = l i m x → 0 − l o g ( 1 + x a 1 − x b ) x \mathrm{LHL}=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{\mathrm{log}(\frac{1+\frac{x}{a}}{1-\frac{x}{b}})}{x} LHL = x → 0 − lim x log ( 1 − b x 1 + a x )
= l i m x → 0 − l o g ( 1 + x a ) − l o g ( 1 − x b ) x =\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{\mathrm{log}(1+\frac{x}{a})-\mathrm{log}(1-\frac{x}{b})}{x} = x → 0 − lim x log ( 1 + a x ) − log ( 1 − b x )
= l i m x → 0 − l o g ( 1 + x a ) x a × 1 a − l o g ( 1 − x b ) − x b × − 1 b = 1 b + 1 a =\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{\mathrm{log}(1+\frac{x}{a})}{\frac{x}{a}}\times \frac{1}{a}-\frac{\mathrm{log}(1-\frac{x}{b})}{-\frac{x}{b}}\times -\frac{1}{b}=\frac{1}{b}+\frac{1}{a} = x → 0 − lim a x log ( 1 + a x ) × a 1 − − b x log ( 1 − b x ) × − b 1 = b 1 + a 1
R H L = l i m x → 0 + c o s 2 x − s i n 2 x − 1 x 2 + 1 − 1 × x 2 + 1 + 1 x 2 + 1 + 1 \mathrm{RHL}=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x-1}{\sqrt{{x}^{2}+1}-1}\times \frac{\sqrt{{x}^{2}+1}+1}{\sqrt{{x}^{2}+1}+1} RHL = x → 0 + lim x 2 + 1 − 1 cos 2 x − sin 2 x − 1 × x 2 + 1 + 1 x 2 + 1 + 1
= l i m x → 0 + − 2 s i n 2 x x 2 ( x 2 + 1 + 1 ) = − 4 =\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}-\frac{2{\mathrm{sin}}^{2}x}{{x}^{2}}(\sqrt{{x}^{2}+1}+1)=-4 = x → 0 + lim − x 2 2 sin 2 x ( x 2 + 1 + 1 ) = − 4
So, 1 b + 1 a = k = − 4. \frac{1}{b}+\frac{1}{a}=k=-4. b 1 + a 1 = k = − 4.
k = − 4 ⇒ 4 k = − 1. k=-4\Rightarrow \frac{4}{k}=-1. k = − 4 ⇒ k 4 = − 1.
1 a + 1 b = − 4. \frac{1}{a}+\frac{1}{b}=-4. a 1 + b 1 = − 4.
∴ 1 a + 1 b + 4 k = − 5. \therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=-5. ∴ a 1 + b 1 + k 4 = − 5.