Question

If $\underset{x\to 0}{\lim }{\left(1+px+q{x}^2\right)}^{cosecx}=2048$ , then the value of $\frac{p}{11}$ is equal to (take $ln2=0.69$ )

Answer 1

Answer: 0.69

The given limit is of the form $1^{\in fty}$
$\therefore e^{\underset{x\to 0}{\lim }\left(cosecx\right)\left(1+px+q{x}^2-1\right)}=2048$
$\underset{x\to 0}{\lim }\left(\frac{px+q{x}^2}{sinx}\right)=ln2048$
$\underset{x\to 0}{\lim }\left(\frac{px+q{x}^2}{x}\right)\cdot \left(\frac{x}{sinx}\right)=11ln2$
$\underset{x\to 0}{\lim }\left(p+qx\right)\left(1\right)=11ln2\Rightarrow p=11ln2$
Hence, $\frac{p}{11}=ln2=0.69$