Question

The value of the expression $\underset{x\to \infty }{lim}(1+t^2{)}^x\sin \frac{k}{(1+t^2{)}^x}$ is $(t,k\in R)$ is

• A. $k^{1/2}$
• B. $k$
• C. $k^2$
• D. $k^{1/3}$

Answer 1

Answer: (B)

$k$

Given $li{m}_{x\to \infty }(1+t^2{)}^x\sin \frac{k}{(1+t^2{)}^x}$

$=li{m}_{x\to \infty }\frac{\sin \frac{k}{(1+t^2{)}^x}}{\frac{1}{(1+t^2{)}^x}}$
As $x$ tends to $\infty$, $\frac{k}{(1+t^2{)}^x}$ tends to 0

multiply and divide by $k$
$li{m}_{x\to \infty }\frac{k\sin \left(\frac{k}{(1+t^2{)}^x}\right)}{\frac{k}{(1+t^2{)}^x}}$ $(\because {lim}_{x\to 0}\frac{\sin x}{x}=1)$
$=k$