Functions Question 95
Question: $ \underset{y\to 0}{\mathop{\lim }},\frac{(x+y)\sec (x+y)-x\sec x}{y}= $
[AI CBSE 1990]
Options:
A) $ \sec x(x\tan x+1) $
B) $ x\tan x+\sec x $
C) $ x\sec x+\tan x $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{y\to 0}{\mathop{\lim }}{ \frac{x,{ \sec ,(x+y)-\sec x }}{y}+\sec ,(x+y) } $ $ =\underset{y\to 0}{\mathop{\lim }}[ \frac{x}{y},{ \frac{\cos x-\cos ,(x+y)}{\cos ,(x+y),\cos x} } ]+\underset{y\to 0}{\mathop{\lim }},\sec ,(x+y) $ $ =\underset{y\to 0}{\mathop{\lim }}[ \frac{x\sin ,( x+\frac{y}{2} )}{\cos ,(x+y),.\cos x},.,\frac{\sin ,( \frac{y}{2} )}{,\frac{y}{2}} ]+\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) $
BETA
