Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{\log(2+x)-\log(2-x)}{x}$

Answer 1

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Answer 2

1st method : L-Hospital rule,

you can easily solve above limit using L-HOSPITAL rule

as given, limit is in the form of 0/0

differentiating numerator and denominator individually,

i.e., $\displaystyle\lim_{x\to 0}\frac{\frac{1}{(2+x)}-\frac{-1}{(2-x)}}{1}$

now putting, x = 0

so, 1/(2 + 0) + 1/(2 - 0) = 1/2 + 1/2 = 1

hence, answer is 1

method 2 : in this method , you have to standard form of limit.

i.e., $\displaystyle\lim_{x\to 0}\frac{log(1+f(x))}{f(x)}=1$ where f(x) = 0 at x = 0

so, $\displaystyle\lim_{x\to 0}\frac{log2(1+x/2)-log2(1-x/2)}{x}$

= $\displaystyle\lim_{x\to 0}\frac{log2 + log(1 + x/2) - log2 - log(1 - x/2)}{x}$

= $\displaystyle\lim_{x\to 0}\frac{log(1+x/2)-log(1-x/2)}{x}$

= $\displaystyle\lim_{x\to 0}\frac{log(1+x/2)}{x/2}\times\frac{1}{2}-\displaystyle\lim{x\to 0}\frac{log(1-x/2)}{-x/2}\times\frac{-1}{2}$

= 1 × 1/2 - 1 × (-1/2)

= 1/2 + 1/2

= 1 [Ans]