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find the limit.
\\(\lim_{x\to0}\frac{\sin(7x)}{x}\\)
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  1. -/1 points

find the limit.

Explanation:

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

ightarrow0}\frac{\sin u}{u}=1$
Let $u = 7x$. As $x
ightarrow0$, then $u
ightarrow0$. And $\frac{\sin(7x)}{x}=7\times\frac{\sin(7x)}{7x}$.

Step2: Evaluate the limit

$\lim_{x
ightarrow0}\frac{\sin(7x)}{x}=\lim_{x
ightarrow0}7\times\frac{\sin(7x)}{7x}$. Since $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$ and here $u = 7x$, we have $\lim_{x
ightarrow0}7\times\frac{\sin(7x)}{7x}=7\times\lim_{7x
ightarrow0}\frac{\sin(7x)}{7x}$.

Answer:

$7$