Question

Use the series representation of the function \(f\) to find \(\lim _{x \rightarrow 0} f(x)\) (if it exists). $$ f(x)=\frac{\sin x}{x} $$

Short answer

The limit as \(x\) approaches 0 of \(f(x)=\frac{\sin x}{x}\) is 1.

Step-by-step solution

Substituting the Maclaurin series representation for sin(x)

Let's substitute the series representation of \(\sin x\) into the function. We get: \( f(x) = \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots\)

Find the limit

Now, let's take the limit as \(x\) approaches 0: \( \lim_{x \rightarrow 0} f(x) =\lim_{x \rightarrow 0} (1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots)\)

Substitute x=0 into the function

If we substitute \(x=0\) into the function we get: \( \lim_{x \rightarrow 0} f(x) = 1 - \frac{0}{3!} + \frac{0}{5!} - \frac{0}{7!} + \cdots = 1 \).