Question

use lhôpitals rule to evaluate $lim_{x \to 0}\frac{sin 13x}{x}$. then determine the limit using limit laws and commonly known limits. use lhôpitals rule to rewrite the given limit so that it is not an indeterminate form. $lim_{x \to 0}\frac{sin 13x}{x}=lim_{x \to 0}square$

Explanation:

Step1: Check indeterminate form

As $x\to0$, $\lim_{x\to0}\frac{\sin(13x)}{x}$ is in the $\frac{0}{0}$ form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $\sin(13x)$ with respect to $x$ is $13\cos(13x)$ (by the chain - rule, if $y = \sin(u)$ and $u = 13x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\cos(u)\cdot13 = 13\cos(13x)$), and the derivative of $x$ with respect to $x$ is $1$. So $\lim_{x\to0}\frac{\sin(13x)}{x}=\lim_{x\to0}\frac{13\cos(13x)}{1}$.

Step3: Evaluate the limit

Substitute $x = 0$ into $\frac{13\cos(13x)}{1}$. We know that $\cos(0)=1$, so $\lim_{x\to0}\frac{13\cos(13x)}{1}=13\cos(0)=13$.

Answer:

$13$