Chapter 11.3, Problem 50E

Solution

(a)

To find: The value of the limit $\underset{x\to \infty }{{\displaystyle \lim y}}$ for the given function $y=\frac{x}{1+2^x}$ .

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

Given information:

The given function is $y=\frac{x}{1+2^x}$ .

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

L’Hospital’s Rule:

If $\underset{x\to c}{{\displaystyle \lim f\left(x\right)}}=\underset{x\to c}{{\displaystyle \lim g\left(x\right)=0/\pm \infty }}$ and $g^{\prime }(x)\ne 0$ for all $x\in I$ with $x\ne c$ and $\frac{\underset{x\to c}{{\displaystyle \lim f^{\prime }\left(x\right)}}}{\underset{x\to c}{{\displaystyle \lim g^{\prime }\left(x\right)}}}$ exists, then

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

Calculation:

The given function is: $y=\frac{x}{1+2^x}$

Substitute $x=\infty$ in the denominator

  $2^{\infty }$ means $2\times 2\times 2\times \mathrm{2......}$ up to infinity which tends to infinity.

i.e. $\underset{x\to \infty }{{\displaystyle \lim 2^x=\infty }}$

Calculate the limits:

  $\underset{x\to \infty }{{\displaystyle \lim y}}=\underset{x\to \infty }{{\displaystyle \lim }}\left(1+2^x\right)$

According to Sum rule of limits:

  $\underset{}{{\displaystyle \underset{x\to \infty }{{\displaystyle \lim y}}=\underset{x\to \infty }{{\displaystyle \lim 1}}+\underset{x\to \infty }{{\displaystyle \lim 2^x}}}}$

Substitute $\underset{x\to \infty }{{\displaystyle \lim 2^x=\infty }}$ and $\underset{x\to \infty }{{\displaystyle \lim 1=1}}$

in the above equation to obtain the required value of the limit:

  $\underset{}{{\displaystyle \begin{array}{l}\underset{x\to \infty }{{\displaystyle \lim \left(1+2^x\right)}}=1+\infty \\ \underset{x\to \infty }{{\displaystyle \lim \left(1+2^x\right)}}=\infty \end{array}}}$

Therefore,

  $\begin{array}{l}\underset{x\to \infty }{{\displaystyle \lim y}}=\underset{x\to \infty }{\lim }\frac{x}{1+2^x}\\ \underset{x\to \infty }{{\displaystyle \lim y}}=\frac{\infty }{\infty }\end{array}$

Since, $\underset{x\to \infty }{{\displaystyle \lim y}}=\frac{\infty }{\infty }$

Apply L’Hospital’s rule to obtain the required limit:

  $\begin{array}{l}\underset{x\to \infty }{{\displaystyle \lim y=}}\underset{x\to \infty }{\lim }\frac{x}{1+2^x}\\ \underset{x\to \infty }{{\displaystyle \lim y}}=\frac{\underset{x\to \infty }{{\displaystyle \lim }}\frac{d}{dx}\left(x\right)}{\underset{x\to \infty }{{\displaystyle \lim }}\frac{d}{dx}\left(1+2^x\right)}\\ \underset{x\to \infty }{{\displaystyle \lim y}}=\frac{\underset{x\to \infty }{{\displaystyle \lim }}1}{\underset{x\to \infty }{{\displaystyle \lim }}{2}^x\ln 2}\\ \underset{x\to \infty }{{\displaystyle \lim y}}=\frac{1}{\infty }\\ \underset{x\to \infty }{{\displaystyle \lim y}}=0\end{array}$

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

(b)

To find: The value of the limit $\underset{x\to -\infty }{{\displaystyle \lim y}}$ for the given function $y=\frac{x}{1+2^x}$ .

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

Given information:

The given function is $y=\frac{x}{1+2^x}$ .

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

Quotient 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 $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to \infty }{{\displaystyle \lim f(x)}}}{\underset{x\to \infty }{{\displaystyle \lim g(x)}}}$ .

Calculation:

The given function is: $y=\frac{x}{1+2^x}$

Substitute $x=-\infty$

  $2^{-\infty }$ means $\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \mathrm{......}$ up to infinity which tends to $0$ .

i.e. $\underset{x\to -\infty }{{\displaystyle \lim 2^x=0}}$

Calculate the limits:

  $\underset{x\to -\infty }{{\displaystyle \lim y}}=\underset{x\to -\infty }{{\displaystyle \lim }}\left(1+2^x\right)$

According to Sum rule of limits:

  $\underset{}{{\displaystyle \underset{x\to -\infty }{{\displaystyle \lim \left(1+2^x\right)}}=\underset{x\to -\infty }{{\displaystyle \lim 1}}+\underset{x\to -\infty }{{\displaystyle \lim 2^x}}}}$

Substitute $\underset{x\to -\infty }{{\displaystyle \lim 2^x=0}}$ and $\underset{x\to -\infty }{{\displaystyle \lim 1=1}}$

in the above equation to obtain the required value of the limit:

  $\underset{}{{\displaystyle \begin{array}{l}\underset{x\to -\infty }{{\displaystyle \lim \left(1+2^x\right)}}=1+0\\ \underset{x\to -\infty }{{\displaystyle \lim \left(1+2^x\right)}}=1\end{array}}}$

Calculate the required limit:

  $\underset{x\to -\infty }{{\displaystyle \lim y=}}\underset{x\to -\infty }{\lim }\frac{x}{1+2^x}$

Apply Quotient rule of limits:

  $\underset{x\to -\infty }{{\displaystyle \lim y=}}\frac{\underset{x\to -\infty }{{\displaystyle \lim x}}}{\underset{x\to -\infty }{{\displaystyle \lim (1+2^x)}}}$

Substitute $\underset{x\to -\infty }{{\displaystyle \lim \left(1+2^x\right)}}=1$ and $\underset{x\to -\infty }{\lim x}=-\infty$

in the above equation to obtain the required value:

  $\begin{array}{l}\underset{x\to -\infty }{{\displaystyle \lim y=}}\frac{-\infty }{1}\\ \underset{x\to -\infty }{{\displaystyle \lim y=}}-\infty \end{array}$

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