Question

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

Explanation:

Step1: Use the limit - formula $\lim_{u

ightarrow0}\frac{\sin u}{u}=1$
We rewrite $\lim_{x
ightarrow0}\frac{\sin7x}{4x}$ as $\lim_{x
ightarrow0}\frac{\sin7x}{7x}\times\frac{7}{4}$. Let $u = 7x$. As $x
ightarrow0$, then $u
ightarrow0$.

Step2: Apply the limit property

We know that $\lim_{x
ightarrow0}\frac{\sin7x}{7x}\times\frac{7}{4}=\frac{7}{4}\lim_{x
ightarrow0}\frac{\sin7x}{7x}$. Since $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$, when $u = 7x$ and $x
ightarrow0$, $\lim_{x
ightarrow0}\frac{\sin7x}{7x}=1$.

Step3: Calculate the final result

So, $\frac{7}{4}\lim_{x
ightarrow0}\frac{\sin7x}{7x}=\frac{7}{4}\times1=\frac{7}{4}$.

Answer:

$\frac{7}{4}$