Question
Limitx--->0 sin7x/5x
Answers
Answered by
GPT-4o mini
To evaluate the limit
\[
\lim_{x \to 0} \frac{\sin(7x)}{5x},
\]
we can use the fact that \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\).
First, we rewrite the limit in terms of \(u\):
Let \(u = 7x\). Then as \(x \to 0\), \(u \to 0\) as well. Therefore, we can express \(x\) in terms of \(u\):
\[
x = \frac{u}{7}.
\]
Now, substituting this into the limit gives:
\[
\lim_{x \to 0} \frac{\sin(7x)}{5x} = \lim_{u \to 0} \frac{\sin(u)}{5 \cdot \frac{u}{7}} = \lim_{u \to 0} \frac{\sin(u)}{\frac{5u}{7}} = \lim_{u \to 0} \frac{7 \sin(u)}{5u}.
\]
Now, we can separate the limit:
\[
= \frac{7}{5} \lim_{u \to 0} \frac{\sin(u)}{u} = \frac{7}{5} \cdot 1 = \frac{7}{5}.
\]
Thus, the limit is
\[
\boxed{\frac{7}{5}}.
\]
\[
\lim_{x \to 0} \frac{\sin(7x)}{5x},
\]
we can use the fact that \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\).
First, we rewrite the limit in terms of \(u\):
Let \(u = 7x\). Then as \(x \to 0\), \(u \to 0\) as well. Therefore, we can express \(x\) in terms of \(u\):
\[
x = \frac{u}{7}.
\]
Now, substituting this into the limit gives:
\[
\lim_{x \to 0} \frac{\sin(7x)}{5x} = \lim_{u \to 0} \frac{\sin(u)}{5 \cdot \frac{u}{7}} = \lim_{u \to 0} \frac{\sin(u)}{\frac{5u}{7}} = \lim_{u \to 0} \frac{7 \sin(u)}{5u}.
\]
Now, we can separate the limit:
\[
= \frac{7}{5} \lim_{u \to 0} \frac{\sin(u)}{u} = \frac{7}{5} \cdot 1 = \frac{7}{5}.
\]
Thus, the limit is
\[
\boxed{\frac{7}{5}}.
\]