GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 17

Answer 1

$\begin{align*} {\lim_{x\rightarrow \infty}(x+sinx)^1/x} &= \lim_{x \rightarrow \infty}e^{ln(x+sinx)^{1/x}} \\ &= \lim_{x \rightarrow \infty}e^{\frac{ln(x+sinx)}{x}} \end{align*}$

$e^{x}$ is continous for all $x$

We can take limit inside

$e^{\lim_{x \rightarrow \infty} \frac{ln(x+sinx)}{x}}$

Consider only $\lim_{x \rightarrow \infty} \frac{ln(x+sinx)}{x}$

This is in $\frac{\infty}{\infty}$ form

Apply L’Hopital’s Rule

$\lim_{x \rightarrow \infty} \frac{ln(x+sinx)}{x} = 0 $

Finally

$\lim_{x\rightarrow \infty}$ $(x+sinx)^{1/x} = e^{0}$ =$1$

Comments

Small change required, Just because $e^x$ is continuous for all x, we can't the limit the inside of it, the actual reason is if the limit exists as $e^x$ is continuous for all x, so the composition value at the point will be same as taking the limit value to the power. so it's more like  $lim_{x\to c} f(x) = b$ and $g(x)$ is continuous at $x=b$ then limit value of g(f(x)) as x approaches c will be g(b).

Answer 2

$\lim_{x\to\infty}(x+sinx)^{\frac{1}{x}}$, this is in form of $\infty^{0}$, so we have to do algebraic manipulations to get into either $\frac{\infty}{\infty}$ or $\frac{0}{0}$

Let $ y = \lim_{x\to\infty}(x+sinx)^{\frac{1}{x}}$

$\ln(y) = \lim_{x\to\infty}\frac{1}{x}\ln(x + sinx) = \dfrac{\ln(x+sinx)}{x}$

Now let’s evaluate  $\ln(y)$, so that if we get any finite value we take it as the power of $e$.

$\lim_{x\to\infty}\dfrac{ln(x+sinx)}{x}$  (this is in form of $\frac{\infty}{\infty}$, so apply L’Hopital Rule.)

-= $\lim_{x\to\infty}\dfrac{1}{x+sinx}(1+cosx)$

 =$\lim_{x\to\infty}\dfrac{1+cosx}{x+sinx}$

 

 = $\lim_{x\to\infty}\dfrac{1+cosx}{x(1+\frac{sinx}{x})}$

 

 =$\lim_{x\to\infty}\dfrac{1+cosx}{x}.\dfrac{1}{(1+\frac{sinx}{x})}$

 

  =$\lim_{x\to\infty}\left(\dfrac{1}{x}+\dfrac{cosx}{x}\right)\left(\dfrac{1}{1+\frac{sinx}{x}}\right)$

 

To evaluate $\dfrac{sinx}{x}, \dfrac{cosx}{x}$, we can use the squeeze theorem which says if the function $g(x)$ is in between $f(x), h(x)$ and limit values of $f(x),h(x)$ are equal then limit value of $g(x)$ also be the same.

here,

 

$-1 \leq sinx \leq 1 $ (this is true for every x)    ,    $-1 \leq cosx \leq 1 $ (this is true for every x)

 

$\dfrac{-1}{x} \leq \dfrac{sinx}{x} \leq \dfrac{1}{x}$                                                 $\dfrac{-1}{x} \leq \dfrac{cos}{x} \leq \dfrac{1}{x}$

 

we all know that, $\lim_{x\to\infty}\dfrac{1}{x}=0 $ and $\lim_{x\to\infty}\dfrac{-1}{x}=0 $, so by squeeze theorem $\lim_{x\to\infty}\dfrac{sinx}{x}=0 $ and $\lim_{x\to\infty}\dfrac{cosx}{x}=0 $

 

now substitute these values in the above expression,

 

=$\lim_{x\to\infty}\left(\dfrac{1}{x}+\dfrac{cosx}{x}\right)\left(\dfrac{1}{1+\frac{sinx}{x}}\right)$ = $\lim_{x\to\infty}\left(0+0\right)\left(\dfrac{1}{1+0}\right)$  = $0 (1)$ = 0

Don’t forget that this is the value of $\ln(y)$, to the get value of y take $e$ on both sides, so the final answer will be $e^0$ = 1.