Let f (x)=(ln(1+ax)-ln(1-bx)/x), x ≠ 0. If f(x) is continuous at x = 0, then f(0) =

Question

Let f (x)=(ln(1+ax)-ln(1-bx)/x), x ≠ 0. If f(x) is continuous at x = 0, then f(0) = (A) a - b (B) a + b (C) b - a (D) ln a + ln b

Tardigrade Admin · Accepted Answer

displaystyle limx → 0 f (x) = displaystyle limx → 0 (ln(1+ax)-ln(1-bx)/x) ((0/0) form) = displaystyle limx → 0 [(a/1+ax)+(b/1-bx)] (L' Hospital Rule) =a+b.

Q. Let $f (x) = \frac{l n ( 1 + a x ) - l n ( 1 - b x )}{x}$ , $x \neq = 0$ .
If $f (x)$ is continuous at $x = 0$ , then $f (0) =$

$a - b$

$a + b$

$b - a$

$l n a + l n b$

$x \to 0 lim f (x)$
$= x \to 0 lim$ $\frac{l n ( 1 + a x ) - l n ( 1 - b x )}{x}$ $(\frac{0}{0}$ form $)$
$= x \to 0 lim$ $[\frac{a}{1 + a x} + \frac{b}{1 - b x}]$ ( $L^{'}$ Hospital Rule)
$= a + b$ .

Q. Let f(x)=xln(1+ax)−ln(1−bx)​, x=0. If f(x) is continuous at x=0, then f(0)=

a−b

a+b

b−a

lna+lnb