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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits1 Mark
MCQ
$\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right]=$
  • A
    $\frac{3}{2}$
  • B
    $-\frac{5}{2}$
  • C
    $-\frac{1}{2}$
  • ✓
    $\frac{2}{5}$
✓

Answer

Correct option: D.
$\frac{2}{5}$
(D) $\frac{2}{5}$
Hint:
$\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right] $
$=\lim _{x \rightarrow 0} \frac{\log \left[5\left(1+\frac{x}{5}\right)\right]-\log \left[5\left(1-\frac{x}{5}\right)\right]}{\sin x} $
$=\lim _{x \rightarrow 0} \frac{\log 5+\log \left(1+\frac{x}{5}\right)-\left[\log 5+\log \left(1-\frac{x}{5}\right)\right]}{\sin x} $
$=\lim _{x \rightarrow 0}\left[\frac{\log \left(1+\frac{x}{5}\right)-\log \left(1-\frac{x}{5}\right)}{x} \times \frac{x}{\sin x}\right] $
$=\lim _{x \rightarrow 0}\left[\frac{\log \left(1+\frac{x}{5}\right)}{5\left(\frac{x}{5}\right)}-\frac{\log \left(1-\frac{x}{5}\right)}{(-5)\left(\frac{-x}{5}\right)}\right] \times \lim _{x \rightarrow 0} \frac{x}{\sin x} $
$=\left[\frac{1}{5}(1)+\frac{1}{5}(1)\right] \times 1=\frac{2}{5}$

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