Chapter 13.1, Problem 22E

To determine

To Get: The limit value of given expression at x tends to $1$

Answer to Problem 22E

The limit value of the expression at x tends to $1$ is $0\cdot 23$

Explanation of Solution

Given information:

Function expression $\frac{f(x)}{g(x)}$ is given at $x\to 0$

Concept used:

Use the L’ Hospital rule to evaluate the indeterminate limits.

  $\underset{x\to d}{\lim }\frac{f(x)}{g(x)}=\underset{x\to d}{\lim }\frac{f\text{'}(x)}{g\text{'}(x)}$

By observing $f(x)=x^2$ and $g(x)$ is $\cos 5x-\cos 4x$

First substitute $x=0$ in the denominator. If denominator value is $0$ then use the L’ Hospital Rule to evaluate.

  $\begin{array}{l}\Rightarrow \underset{x\to 0}{\lim }\frac{x^2}{\cos 5x-\cos 4x}\\ \Rightarrow \underset{x\to 0}{\lim }\frac{2x}{-5\sin 5x+4\sin 4x}\left(\text{By using L'Hospital's Rule}\right)\\ \Rightarrow \underset{x\to 0}{\lim }\frac{2}{-25\cos 5x+16\text{cos}4x}\left(\text{By using L'Hospital's Rule}\right)\\ \Rightarrow \frac{2}{-25+16}=\frac{-2}{9}\\ \Rightarrow 0\cdot 23\end{array}$

Hence,

The limit value of the expression at x tends to $1$ is $0\cdot 23$