A) \[\sec x(x\tan x+1)\]
B) \[x\tan x+\sec x\]
C) \[x\sec x+\tan x\]
D) None of these
Correct Answer: A
Solution :
\[\underset{y\to 0}{\mathop{\lim }}\,\,\left\{ \frac{x\,\left\{ \sec \,(x+y)-\sec x \right\}}{y}+\sec \,(x+y) \right\}\] \[=\underset{y\to 0}{\mathop{\lim }}\,\,\left[ \frac{x}{y}\,\left\{ \frac{\cos x-\cos \,(x+y)}{\cos \,(x+y)\,\cos x} \right\} \right]+\underset{y\to 0}{\mathop{\lim }}\,\sec \,(x+y)\] \[=\underset{y\to 0}{\mathop{\lim }}\,\,\left[ \frac{x\sin \,\left( x+\frac{y}{2} \right)}{\cos \,(x+y)\,.\,\,\cos x}\,.\,\frac{\sin \,\left( \frac{y}{2} \right)}{\,\,\,\frac{y}{2}} \right]+\sec x\] = xtanxsecx + secx = secx(xtanx+1). Aliter : Apply L-Hospital?s rule, \[\underset{y\to 0}{\mathop{\lim }}\,\,\frac{(x+y)\,\sec \,(x+y)-x\,\sec x}{y}\] \[=\underset{y\to 0}{\mathop{\lim }}\,\,\frac{(x+y)\,\sec \,(x+y)\tan \,(x+y)+\sec \,(x+y)-0}{1}\] {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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