Question

The value of $\underset{x\to \frac{\pi }{2}}{lim}{(2^{\frac{1}{{\cos }^{2}x}}+3^{\frac{1}{{\cos }^{2}x}}+4^{\frac{1}{{\cos }^{2}x}}+5^{\frac{1}{{\cos }^{2}x}}+6^{\frac{1}{{\cos }^{2}x}})}^{2{\cos }^{2}x}$ is

• A. $1$
• B. $6$
• C. $36$
• D. $\frac{1}{36}$

Answer 1

The correct option is C $36$

Let $\frac{1}{{\cos }^{2}x}=t$
So, $x\to \frac{\pi }{2}\Rightarrow t\to \infty$
We have:
$\underset{t\to \infty }{lim}{(2^t+3^t+4^t+5^t+6^t)}^{\frac{2}{t}}$
$\underset{t\to \infty }{lim}\left(6^2\right)\cdot {({\left(\frac{2}{6}\right)}^t+{\left(\frac{3}{6}\right)}^t+{\left(\frac{4}{6}\right)}^t+{\left(\frac{5}{6}\right)}^t+1)}^{\frac{2}{t}}$
$=36(0+0+0+0+1{)}^0=36$