Chapter 9.2, Problem 5E

To determine

To Find: The value of a limit using L'Hôpital's Rule

Answer to Problem 5E

The value of the limit is $\frac{1}{2}$

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}$

Concept Used:

  $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(a)=g(a)=0,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: At $x=0$ , the value of the expression is $\frac{1-\cos 0}{0^2}=\frac{1-1}{0}=\frac{0}{0}$

So, apply L'Hôpital's Rule to write

  $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\underset{x\to 0}{\lim }\frac{0+\sin x}{2x}$

  $=\frac{1}{2}\underset{x\to 0}{\lim }\frac{\sin x}{x}$

  $=\frac{1}{2}\cdot 1=\frac{1}{2}$

Therefore, the limit is $\frac{1}{2}$