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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
MCQ
$\lim _{x \rightarrow 0} \frac{\sin x+\log (1-x)}{x^2}$ is equal to
  • A
    $0$
  • B
    $\frac{1}{2}$
  • ✓
    $-\frac{1}{2}$
  • D
    None of these
✓

Answer

Correct option: C.
$-\frac{1}{2}$
(C)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\cos x-\frac{1}{1-x}}{2 x}=\lim _{x \rightarrow 0} \frac{-\sin x-\frac{1}{(1-x)^2}}{2}=-\frac{1}{2}$
Alternate method:
$\lim _{x \rightarrow 0} \frac{\sin x+\log (1-x)}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots\right)}{x^2}$ $+\lim _{x \rightarrow 0} \frac{\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots\right)}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\frac{-x^2}{2}-x^3\left(\frac{1}{3!}+\frac{1}{3}\right)-\frac{x^4}{4} \ldots}{x^2}=-\frac{1}{2}$

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