Question

In Exercises 49 and \(50,\) use the series representation of the function \(f\) to find \(\lim _{x \rightarrow 0} f(x)\) (if it exists). $$ f(x)=\frac{1-\cos x}{x} $$

Short answer

The value of \(\lim _{x \rightarrow 0} f(x)\) is 0.

Step-by-step solution

Express the function in terms of limits

The function is \(f(x) = \frac{1 - cos(x)}{x}\), which, as \(x \rightarrow 0\), takes the form \(0/0\). This indicates an indeterminate form, and hence, l'Hopital's rule can be applied. So, find the limit as \(x \rightarrow 0\) : \(\lim _{x \rightarrow 0} \frac{1 - cos(x)}{x}\)

Apply l'Hopital's rule

According to l'Hopital's rule, the limit of a fraction where both the numerator and the denominator approach zero can be found by taking the derivative of the numerator and the denominator separately. So, differentiating \(1 - cos(x)\) with respect to \(x\) gives \(sin(x)\) and differentiating \(x\) with respect to \(x\) gives \(1\). Now, put these values back into the limit: \(\lim _{x \rightarrow 0} \frac{sin(x)}{1}\)

Simplify the function and find the limit

With the new function \(\lim _{x \rightarrow 0} sin(x)\), which doesn't have an indeterminate form, you can directly substitute the value of \(x = 0\) into the function. \(sin(0) = 0\). Hence, \(\lim _{x \rightarrow 0} sin(x) = 0\)