Question

If $\underset{x \rightarrow 0}{lim} \left(1 + p x + q x^{2}\right)^{c o s e c \, x}=2048$ , then the value of $\frac{p}{11}$ is equal to (take $ln 2=0.69$ )

Answer 1

The given limit is of the form $1^{\in fty}$
$\therefore e^{\underset{x \rightarrow 0}{lim} \left(c o s e c \, x\right) \left(1 + p x + q x^{2} - 1\right)}=2048$
$\underset{x \rightarrow 0}{lim} \left(\frac{p x + q x^{2}}{sin ⁡ x}\right)=ln ⁡ 2048$
$\underset{x \rightarrow 0}{lim} \left(\frac{p x + q x^{2}}{x}\right)\cdot \left(\frac{x}{sin ⁡ x}\right)=11ln ⁡ 2$
$\underset{x \rightarrow 0}{lim} \left(p + q x\right)\left(1\right)=11ln⁡2\Rightarrow p=11ln⁡2$
Hence, $\frac{p}{11}=ln2=0.69$