Chapter 9, Problem 35RE

a.

To determine

To show: The given function has removable discontinuity.

Answer to Problem 35RE

As for the value $x=0$ , both numerator as well as denominator vanishes, then the given function has removable discontinuity at $x=0$ .

Explanation of Solution

Given information : Given function is $f\left(x\right)=\frac{2^{\sin x}-1}{e^x-1}$

Concept used : $f$ has removable discontinuity at $x=0$ if both numerator and denominator of a function are $0$

Calculation :

As for the value $x=0$ , both numerator as well as denominator vanishes, then the given function has removable discontinuity at $x=0$ .

b.

To determine

To define: $f$ at $x=0$

Answer to Problem 35RE

Explanation of Solution

Given information : Given function is $f\left(x\right)=\frac{2^{\sin x}-1}{e^x-1}$

Formula used :

L’Hospital Rule:

  ${\displaystyle \underset{a}{\overset{b}{\int }}\frac{dt}{t}}=\left[\ln b-\ln a\right]$

Calculation:

Now,

  $\begin{array}{l}\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{2^{\sin x}-1}{e^x-1}\left[\frac{0}{0}\text{ form}\right]\\ \text{using L'Hospital rule we get, }\underset{x\to 0}{\lim }\frac{2^{\sin x}\ln \left(2\right)\cos x}{e^x}=\frac{0}{1}=0\\ \therefore \text{ The function can be expressed as }f\left(x\right)=\{\begin{array}{l}\frac{2^{\sin x}-1}{e^x-1}x\ne 0\\ 0x=0\end{array}\end{array}$