MCQ
$\mathop {\lim }\limits_{y \to 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
(a) $\mathop {\lim }\limits_{y \to 0} \,\left\{ {\frac{{x\,\left\{ {\sec \,(x + y) - \sec x} \right\}}}{y} + \sec \,(x + y)} \right\}$

$ = \mathop {\lim }\limits_{y \to 0} \,\left[ {\frac{x}{y}\,\left\{ {\frac{{\cos x - \cos \,(x + y)}}{{\cos \,(x + y)\,\cos x}}} \right\}} \right] + \mathop {\lim }\limits_{y \to 0} \sec \,(x + y)$

$ = \mathop {\lim }\limits_{y \to 0} \,\left[ {\frac{{x\sin \,\left( {x + \frac{y}{2}} \right)}}{{\cos \,(x + y)\,.\,\,\cos x}}\,.\,\frac{{\sin \,\left( {\frac{y}{2}} \right)}}{{\,\,\,\frac{y}{2}}}} \right] + \sec x$

$= x\ tan\ x\ sec\ x + sec\ x = sec\ x\ (x\ tan\ x+1).$

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{y \to 0} \,\frac{{(x + y)\,\sec \,(x + y) - x\,\sec x}}{y}$

$ = \mathop {\lim }\limits_{y \to 0} \,\frac{{(x + y)\,\sec \,(x + y)\tan \,(x + y) + \sec \,(x + y) - 0}}{1}$

{Differentiating w.r.t.$y$ assuming $x$ as constant}

$ = x\sec x\tan x + \sec x. = \sec x(x\tan x + 1)$

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