Question

If $A=\underset{x\to -2}{lim}\frac{\tan \pi x}{x+2}+\underset{x\to \infty }{lim}{(1+\frac{1}{x^2})}^x$, then which of the following does not hold true?

• A. $A>3$
• B. $A>4$ and A is a transcendental number
• C. $A<4$
• D. $A$ is a transcendental number

Answer 1

The correct option is B $A>4$ and A is a transcendental number

$\underset{x\to -2}{lim}\frac{\tan \pi x}{x+2}=\underset{x\to -2}{lim}\frac{\tan \pi \left(x\right)}{x+2}$

$\Rightarrow \underset{x\to -2}{lim}\frac{\tan \pi x}{x+2}=\underset{y\to 0}{lim}\frac{\tan \pi (y-2)}{y}$

$\Rightarrow \underset{x\to -2}{lim}\frac{\tan \pi x}{x+2}=\underset{y\to 0}{lim}\frac{\tan \pi \left(y\right)}{y}$

Multiply and divide by$\pi$, then the limit value of given question is $\pi$

and $\Rightarrow \underset{x\to \infty }{lim}{(1+\frac{1}{x^2})}^{x^2\left(1/x\right)}=\underset{x\to \infty }{lim}{(1+\frac{1}{x^2})}^{x^2{lim}_{x\to \infty }\frac{1}{x}}=e^0=1$
$\therefore A=\underset{x\to -2}{lim}\frac{\tan \pi x}{x+2}+\underset{x\to \infty }{lim}{(1+\frac{1}{x^2})}^x=\pi +1>4$ and is a transcendental number.