Chapter 11.3, Problem 51E

Solution

(a)

To find: The value of the limit $\underset{x\to \infty }{{\displaystyle \lim y}}$ for the given function $y=x+\sin x$ .

The value of the limit is $\underset{x\to \infty }{{\displaystyle \lim y}}=\infty$ .

Given information:

The given function is $y=x+\sin x$ .

Formula used:

The notation $\underset{x\to \infty }{{\displaystyle \lim y}}=l$ means that $y$ gets arbitrarily close to $l$ as $x$ gets arbitrarily large.

Sum rule of limits:

If $\underset{x\to c}{{\displaystyle \lim f\left(x\right)}}$ and $\underset{x\to c}{{\displaystyle \lim g\left(x\right)}}$ both exist, then $\underset{x\to c}{{\displaystyle \lim }}\left[f\left(x\right)+g\left(x\right)\right]=\underset{x\to c}{{\displaystyle \lim }}f\left(x\right)+\underset{x\to c}{{\displaystyle \lim }}g\left(x\right)$ .

Sandwich theorem:

Let $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ for all $x\to a$ . If $\underset{x\to a}{\lim }f\left(x\right)=L=\underset{x\to a}{\lim }h\left(x\right)$ , then

  $\underset{x\to a}{\lim }g\left(x\right)=L$

Calculation:

The given function is: $y=x+\sin x$ .

Since, $0\le \sin x\le 1$ ,

  $-1\le \sin x\le 1$

Therefore,

  $x-1\le x+\sin x\le x+1$

Calculate the limits:

  $\underset{x\to \infty }{{\displaystyle \lim }}\left(x-1\right)=\underset{x\to \infty }{{\displaystyle \lim }}\left(x+1\right)=\infty$

According to the Sandwich theorem

Since, $\underset{x\to \infty }{{\displaystyle \lim }}\left(x-1\right)=\underset{x\to \infty }{{\displaystyle \lim }}\left(x+1\right)=\infty$

  $\begin{array}{l}\underset{x\to \infty }{{\displaystyle \lim }}\left(x+\sin x\right)=\infty \\ \underset{x\to \infty }{{\displaystyle \lim }}y=\infty \end{array}$

Therefore, the required value of the limit is $\infty$ .

(b)

To find: The value of the limit $\underset{x\to -\infty }{{\displaystyle \lim y}}$ for the given function $y=x+\sin x$ .

The value of the limit is $\underset{x\to -\infty }{{\displaystyle \lim y}}=-\infty$ .

Given information:

The given function is $y=x+\sin x$ .

Formula used:

The notation $\underset{x\to \infty }{{\displaystyle \lim y}}=l$ means that $y$ gets arbitrarily close to $l$ as $x$ gets arbitrarily large.

Sum rule of limits:

If $\underset{x\to c}{{\displaystyle \lim f\left(x\right)}}$ and $\underset{x\to c}{{\displaystyle \lim g\left(x\right)}}$ both exist, then $\underset{x\to c}{{\displaystyle \lim }}\left[f\left(x\right)+g\left(x\right)\right]=\underset{x\to c}{{\displaystyle \lim }}f\left(x\right)+\underset{x\to c}{{\displaystyle \lim }}g\left(x\right)$ .

Sandwich theorem:

Let $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ for all $x\to a$ . If $\underset{x\to a}{\lim }f\left(x\right)=L=\underset{x\to a}{\lim }h\left(x\right)$ , then

  $\underset{x\to a}{\lim }g\left(x\right)=L$

Calculation:

The given function is: $y=x+\sin x$

Since, $0\le \sin x\le 1$ ,

  $-1\le \sin x\le 1$

Therefore,

  $x-1\le x+\sin x\le x+1$

Calculate the limits:

  $\underset{x\to -\infty }{{\displaystyle \lim }}\left(x-1\right)=\underset{x\to -\infty }{{\displaystyle \lim }}\left(x+1\right)=-\infty$

According to the Sandwich theorem

Since, $\underset{x\to -\infty }{{\displaystyle \lim }}\left(x-1\right)=\underset{x\to -\infty }{{\displaystyle \lim }}\left(x+1\right)=-\infty$

  $\begin{array}{l}\underset{x\to \infty }{{\displaystyle \lim }}\left(x+\sin x\right)=-\infty \\ \underset{x\to \infty }{{\displaystyle \lim }}y=-\infty \end{array}$

Therefore, the required value of the limit is $-\infty$ .