Question

If $\underset{x\to 0}{lt}(\cos x+3\sin 2x{)}^{5/x}=e^k$, then , k =

• A. 20
• B. 10
• C. 30
• D. 40

Answer 1

The correct option is C 30

When we put $x=0$, the given limit is of the form $1^{\infty }$

For this form,

$\underset{x\to a}{lim}{\left\{f\right(x\left)\right\}}^{g\left(x\right)}=e^{\underset{x\to a}{lim}g\left(x\right)\left\{f\right(x)-1\}}$

$\underset{x\to 0}{lim}{(\cos x+3\sin 2x)}^{\frac{5}{x}}=e^{\underset{x\to 0}{lim}\frac{5}{x}(\cos x+3\sin 2x-1)}$

$=e^{\underset{x\to 0}{lim}\frac{5}{x}(\cos x-1)+\underset{x\to 0}{lim}\frac{5}{x}\left(3\sin 2x\right)}$

$=e^{\underset{x\to 0}{lim}5\left(\frac{\cos x-1}{x}\right)+\underset{2x\to 0}{lim}30\left(\frac{\sin 2x}{2x}\right)}$

$=e^{5\left(0\right)+30\left(1\right)}=e^{30}$

So, $k=30$