Chapter 4, Problem 30RE

To determine

To find: the limit of the function.

Answer to Problem 30RE

The limit of the function $\underset{\text{x}\to \text{0}}{\text{lim}}\frac{{\text{e}}^{\text{4x}}\text{-1-4x}}{{\text{x}}^{\text{2}}}$ is $8$ .

Explanation of Solution

Given:

  $\underset{\text{x}\to \text{0}}{\text{lim}}\frac{{\text{e}}^{\text{4x}}\text{-1-4x}}{{\text{x}}^{\text{2}}}$ .

Concept used:

If the function will be in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

In this kind of situation $\text{L}$ Hospital’s Rule.

Which state that the limit of a quotient of the functions is equal to the limit of quotient of the derivative.

This rule can be applied as much as necessary so to get not Indeterminate Forms.

Calculation:

  $\underset{\text{x}\to \text{0}}{\text{lim}}\frac{{\text{e}}^{\text{4x}}\text{-1-4x}}{{\text{x}}^{\text{2}}}$

By putting direct $x=0$ the function will be in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

In this kind of situation $\text{L}$ Hospital’s Rule.

Which state that the limit of a quotient of the functions is equal to the limit of quotient of there derivative.

  $\underset{\text{x}\to \text{0}}{\text{lim}}\frac{{\text{e}}^{\text{4x}}\text{-1-4x}}{{\text{x}}^{\text{2}}}$

  $\underset{\text{x}\to \text{0}}{\text{=lim}}\frac{\frac{\text{d}}{\text{dx}}\left({\text{e}}^{\text{4x}}\text{-1-4x}\right)}{\frac{\text{d}}{\text{dx}}\left(x^2\right)}$ .

Finding derivative of numerator and denominator.

  $\underset{\text{x}\to \text{0}}{\text{=lim}}\frac{{\text{4e}}^{\text{4x}}\text{-4}}{2x}$ .

Taking limit of each term:

  $\frac{{\text{4e}}^{\text{4}\left(0\right)}\text{-4}}{2\left(0\right)}=\frac{0}{0}$

Which is in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

This rule can be applied as much as necessary so to get not Indeterminate Forms.

So, differentiating numerator and denominator.

  $\frac{d}{dx}\left(\underset{\text{x}\to \text{0}}{\text{lim}}\frac{{\text{4e}}^{\text{4x}}\text{-4}}{2x}\right)=\underset{\text{x}\to \text{0}}{\text{lim}}\frac{16e^{4x}}{2}=\frac{16}{2}=8$ .

Hence the limit of the function $\underset{\text{x}\to \text{0}}{\text{lim}}\frac{{\text{e}}^{\text{4x}}\text{-1-4x}}{{\text{x}}^{\text{2}}}$ is $8$ .