Chapter 11.3, Problem 53E

Solution

(a)

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

The value of the limit is $\underset{x\to \infty }{{\displaystyle \lim y}}=undefined$ or does not exist.

Given information:

The given function is $y=e^{-x}\sin x$ .

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 $\underset{x\to c}{{\displaystyle \lim }}\left[f\left(x\right)+g\left(x\right)\right]=\underset{x\to c}{{\displaystyle \lim }}f\left(x\right)+\underset{x\to c}{{\displaystyle \lim }}g\left(x\right)$ .

Calculation:

The given function is: $y=e^{-x}\sin x$ .

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

  $\sin x$ oscillates between 0 and 1.

But,

  $e^{-x}$ tends to infinity as $x$ increases arbitrarily.

Therefore, $\underset{x\to \infty }{{\displaystyle \lim y}}$ does not exist.

The graph of $y=e^{-x}\sin x$

confirms this:

  

Figure1

(b)

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

The value of the limit is $\underset{x\to -\infty }{{\displaystyle \lim y}}=0$ .

Given information:

The given function is $y=e^{-x}\sin x$ .

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.

Product 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 $\underset{x\to c}{{\displaystyle \lim }}\left[f\left(x\right)\cdot g\left(x\right)\right]=\underset{x\to c}{{\displaystyle \lim }}f\left(x\right)\cdot \underset{x\to c}{{\displaystyle \lim }}g\left(x\right)$ .

Calculation:

The given function is: $y=e^{-x}\sin x$

  $e^x$ approaches $0$ as $x$ approaches $-\infty$ . So, $e^{-x}=\frac{1}{e^x}$ tends to 0 as $x$ increases arbitrarily towards $-\infty$ .

Therefore, $\underset{x\to -\infty }{{\displaystyle \lim e^{-x}}}=0$

Calculate the limit:

  $\begin{array}{l}\underset{x\to -\infty }{{\displaystyle \lim }}y=\underset{x\to -\infty }{{\displaystyle \lim }}\left(e^{-x}\sin x\right)\\ \underset{x\to -\infty }{{\displaystyle \lim }}y=\underset{x\to -\infty }{{\displaystyle \lim }}{e}^{-x}\cdot \underset{x\to -\infty }{{\displaystyle \lim }}\sin x\\ \underset{x\to -\infty }{{\displaystyle \lim }}y=0.\underset{x\to -\infty }{{\displaystyle \lim }}\sin x\\ \underset{x\to -\infty }{{\displaystyle \lim }}y=0\end{array}$

Therefore, the required value of the limit is $0$ .