Question

If $\underset{x\to 0}{lim}{\left(\cos x+a\sin bx\right)}^{\frac{1}{x}}=e^2$ then
the possible values of ${}^{\prime }{a}^{\prime }{\&}^{\prime }{b}^{\prime }are:$

• A. $a=1,b=2$
• B. $a=2,b=1$
• C. $a=3,b=2/3$
• D. $a=2/3,b=3$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $a=2/3,b=3$
B $a=1,b=2$
C $a=2,b=1$
D $a=3,b=2/3$

$\underset{x\to 0}{lim}{\left(\cos x+a\sin bx\right)}^{\frac{1}{x}}$ is of form $1^{\infty }$, so its limit will be $e^k$, where

$k=\underset{x\to 0}{lim}\frac{1}{x}(cosx+asinbx-1)=\underset{x\to 0}{lim}\frac{-sinx+abcosbx}{1}=ab=2$

Hence all possible combination of $a$ and $b$ are possible whose product is $2$