Condition for continuity at x = 0 x=0 x = 0
For continuity at x = 0 x=0 x = 0
lim x → 0 − f ( x ) = f ( 0 ) = lim x → 0 + f ( x ) = k . \lim_{x\to 0^-} f(x)=f(0)=\lim_{x\to 0^+} f(x)=k. x → 0 − lim f ( x ) = f ( 0 ) = x → 0 + lim f ( x ) = k . So we compute the left-hand and right-hand limits.
Left-hand limit x → 0 − x\to 0^- x → 0 −
Given
f ( x ) = 1 x ln ( 1 + x a 1 − x b ) , x < 0. f(x)=\frac1x\ln\left(\frac{1+\frac xa}{1-\frac xb}\right), \qquad x<0. f ( x ) = x 1 ln ( 1 − b x 1 + a x ) , x < 0. Rewrite:
ln ( 1 + x a 1 − x b ) = ln ( 1 + x a ) − ln ( 1 − x b ) . \ln\left(\frac{1+\frac xa}{1-\frac xb}\right)=\ln\left(1+\frac xa\right)-\ln\left(1-\frac xb\right). ln ( 1 − b x 1 + a x ) = ln ( 1 + a x ) − ln ( 1 − b x ) . Hence
\lim_{x\to 0^-}\frac1x\ln\left(\frac{1+\frac xa}{1-\frac xb}}\right)
=
\lim_{x\to 0}\left[\frac{\ln(1+\frac xa)}x-\frac{\ln(1-\frac xb)}x\right].
Using the standard limit
lim t → 0 ln ( 1 + t ) t = 1 , \lim_{t\to 0}\frac{\ln(1+t)}t=1, t → 0 lim t ln ( 1 + t ) = 1 , we get
lim x → 0 ln ( 1 + x a ) x = 1 a , \lim_{x\to 0}\frac{\ln(1+\frac xa)}x=\frac1a, x → 0 lim x ln ( 1 + a x ) = a 1 , and
lim x → 0 ln ( 1 − x b ) x = lim x → 0 ln ( 1 + ( − x / b ) ) x = − 1 b . \lim_{x\to 0}\frac{\ln(1-\frac xb)}x
=\lim_{x\to 0}\frac{\ln(1+(-x/b))}{x}
=-\frac1b. x → 0 lim x ln ( 1 − b x ) = x → 0 lim x ln ( 1 + ( − x / b )) = − b 1 . Therefore,
lim x → 0 − f ( x ) = 1 a − ( − 1 b ) = 1 a + 1 b . \lim_{x\to 0^-} f(x)=\frac1a-\left(-\frac1b\right)=\frac1a+\frac1b. x → 0 − lim f ( x ) = a 1 − ( − b 1 ) = a 1 + b 1 .
Right-hand limit x → 0 + x\to 0^+ x → 0 +
For x > 0 x>0 x > 0
f ( x ) = cos 2 x − sin 2 x − 1 1 + x 2 − 1 . f(x)=\frac{\cos^2 x-\sin^2 x-1}{\sqrt{1+x^2}-1}. f ( x ) = 1 + x 2 − 1 cos 2 x − sin 2 x − 1 . Use
cos 2 x − sin 2 x = cos 2 x . \cos^2 x-\sin^2 x=\cos 2x. cos 2 x − sin 2 x = cos 2 x . So numerator becomes
cos 2 x − 1 = − 2 sin 2 x . \cos 2x-1=-2\sin^2 x. cos 2 x − 1 = − 2 sin 2 x . Hence
lim x → 0 + f ( x ) = lim x → 0 + − 2 sin 2 x 1 + x 2 − 1 . \lim_{x\to 0^+} f(x)=\lim_{x\to 0^+}\frac{-2\sin^2 x}{\sqrt{1+x^2}-1}. x → 0 + lim f ( x ) = x → 0 + lim 1 + x 2 − 1 − 2 sin 2 x . Rationalize the denominator:
− 2 sin 2 x 1 + x 2 − 1 ⋅ 1 + x 2 + 1 1 + x 2 + 1 = − 2 sin 2 x ( 1 + x 2 + 1 ) x 2 . \frac{-2\sin^2 x}{\sqrt{1+x^2}-1}\cdot \frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}+1}
=
\frac{-2\sin^2 x(\sqrt{1+x^2}+1)}{x^2}. 1 + x 2 − 1 − 2 sin 2 x ⋅ 1 + x 2 + 1 1 + x 2 + 1 = x 2 − 2 sin 2 x ( 1 + x 2 + 1 ) . Thus
lim x → 0 + f ( x ) = − 2 ( lim x → 0 sin x x ) 2 ( lim x → 0 ( 1 + x 2 + 1 ) ) . \lim_{x\to 0^+} f(x)= -2\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2\left(\lim_{x\to 0}(\sqrt{1+x^2}+1)\right). x → 0 + lim f ( x ) = − 2 ( x → 0 lim x sin x ) 2 ( x → 0 lim ( 1 + x 2 + 1 ) ) . Now,
lim x → 0 sin x x = 1 , lim x → 0 ( 1 + x 2 + 1 ) = 2. \lim_{x\to 0}\frac{\sin x}{x}=1,
\qquad
\lim_{x\to 0}(\sqrt{1+x^2}+1)=2. x → 0 lim x sin x = 1 , x → 0 lim ( 1 + x 2 + 1 ) = 2. So
lim x → 0 + f ( x ) = − 2 ⋅ 1 ⋅ 2 = − 4. \lim_{x\to 0^+} f(x)=-2\cdot 1\cdot 2=-4. x → 0 + lim f ( x ) = − 2 ⋅ 1 ⋅ 2 = − 4. Therefore, for continuity,
k = − 4. k=-4. k = − 4.
Equating left and right limits
Since both one-sided limits must equal k k k
1 a + 1 b = k = − 4. \frac1a+\frac1b=k=-4. a 1 + b 1 = k = − 4. Now compute
1 a + 1 b + 4 k = − 4 + 4 − 4 = − 4 − 1 = − 5. \frac1a+\frac1b+\frac4k=-4+\frac4{-4}=-4-1=-5. a 1 + b 1 + k 4 = − 4 + − 4 4 = − 4 − 1 = − 5.
Final answer
− 5 \boxed{-5} − 5 So the correct option is A .