Continuity

Question

If the function f(x) defined by $\left\{\begin{matrix} \frac{log(1+ ax) - log(1-bx)}{x} &, if x \neq 0\\ k & ,if x = 0 \end{matrix}\right.$

is continuous at x = 0, then value of k is

A) b - a

B) a - b

C) a + b

D) -a - b

Comments

yes corrected :P :p

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L hosipital rule can be applied to only 0/0 form . Do you guys know how to covert other form to 0/0 form and solve using l hospital rule

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L hosipital rule can be applied to only 0/0 form  , it can be applied to  infinity /infinity too

Answer 1

condition for continuity = LHL = RHL = f(0)    , here f(0) = k

we have to find limit of this at x=0  $log(1+ax) - log(1-bx) / x$  

so , if i put x=0 i will get indeterminant form of 0/0  L-hospital rule use karo

after differentiating  =$\frac{a}{1+ax} - \frac{-b}{1-bx}$

now put again x=0 , we get a -(-b) = a+b  

LHL = RHL = f(0)

a+b =k

Comments

oh yes L-hospital thanks

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Why we have calculated at x =0 instead of LHS or RHS it should be x-h  ?