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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
MCQ
$\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}=$
  • ✓
    $\sec x(x \tan x+1)$
  • B
    $x \tan x+\sec x$
  • C
    $x \sec x+\tan x$
  • D
    none of these
✓

Answer

Correct option: A.
$\sec x(x \tan x+1)$
(A)
$\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}$
$=\lim _{y \rightarrow 0}\left\{\frac{x[\sec (x+y)-\sec x]}{y}+\sec (x+y)\right\}$
$=\lim _{y \rightarrow 0}\left[\frac{x}{y} \cdot \frac{\cos x-\cos (x+y)}{\cos (x+y) \cos x}\right]+\lim _{y \rightarrow 0} \sec (x+y)$
$=\lim _{y \rightarrow 0}\left\lfloor\frac{x}{y} \cdot \frac{2 \sin \left(x+\frac{y}{2}\right) \sin \left(\frac{y}{2}\right)}{\cos (x+y) \cos x}\right\rfloor+\sec x$
$=\lim _{y \rightarrow 0}\left[\frac{x \sin \left(x+\frac{y}{2}\right)}{\cos (x+y) \cdot \cos x} \cdot \frac{\sin \left(\frac{y}{2}\right)}{\frac{y}{2}}\right]+\sec x$
$=\left(\frac{x \sin x}{\cos x \cdot \cos x} \cdot 1\right)+\sec x$
$=x \tan x \sec x+\sec x=\sec x(x \tan x+1)$

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