Question

evaluate the following limit. use lhôpitals rule when it is convenient and applicable.

$$\lim_{x \to 0} \frac{4 \sin 3x}{7x}$$

use lhôpitals rule to rewrite the given limit so that it is not an indeterminate form.

$$\\lim_{x \\to 0} \\frac{4 \\sin 3x}{7x}=\\lim_{x \\to 0} \\left(\\square\ ight)$$

evaluate the limit.

$$\lim_{x \to 0} \frac{4 \sin 3x}{7x}=\square$$ (type an exact answer.)

Explanation:

Step1: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the indeterminate form \(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\), then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\).
For \(f(x) = 4\sin(3x)\) and \(g(x)=7x\), \(f^{\prime}(x)=4\times3\cos(3x)=12\cos(3x)\) and \(g^{\prime}(x) = 7\).
So, \(\lim_{x
ightarrow0}\frac{4\sin(3x)}{7x}=\lim_{x
ightarrow0}\frac{12\cos(3x)}{7}\).

Step2: Evaluate the limit

Substitute \(x = 0\) into \(\frac{12\cos(3x)}{7}\).
Since \(\cos(0)=1\), we have \(\frac{12\cos(0)}{7}=\frac{12\times1}{7}\).

Answer:

\(\lim_{x
ightarrow0}\frac{4\sin(3x)}{7x}=\frac{12}{7}\)