Question

find the limit.
lim(x→0) sin(7x)/8x

1. -/1 points

find the limit.
lim(t→0) tan(10t)/sin(2t)

Explanation:

Step1: Use the limit formula $\lim_{u

ightarrow0}\frac{\sin u}{u} = 1$
For $\lim_{x
ightarrow0}\frac{\sin(7x)}{8x}$, rewrite it as $\frac{7}{8}\lim_{x
ightarrow0}\frac{\sin(7x)}{7x}$. Let $u = 7x$, as $x
ightarrow0$, $u
ightarrow0$. Then $\lim_{x
ightarrow0}\frac{\sin(7x)}{7x}=\lim_{u
ightarrow0}\frac{\sin u}{u}=1$. So $\frac{7}{8}\lim_{x
ightarrow0}\frac{\sin(7x)}{7x}=\frac{7}{8}\times1=\frac{7}{8}$.

Step2: For $\lim_{t

ightarrow0}\frac{\tan(10t)}{\sin(2t)}$, rewrite $\tan(10t)=\frac{\sin(10t)}{\cos(10t)}$
So $\lim_{t
ightarrow0}\frac{\tan(10t)}{\sin(2t)}=\lim_{t
ightarrow0}\frac{\sin(10t)}{\cos(10t)\sin(2t)}=\lim_{t
ightarrow0}\frac{\sin(10t)}{10t}\times\frac{10t}{2t}\times\frac{2t}{\sin(2t)}\times\frac{1}{\cos(10t)}$.

Step3: Apply the limit formula $\lim_{u

ightarrow0}\frac{\sin u}{u} = 1$
As $t
ightarrow0$, $\lim_{t
ightarrow0}\frac{\sin(10t)}{10t}=1$, $\lim_{t
ightarrow0}\frac{2t}{\sin(2t)} = 1$ and $\lim_{t
ightarrow0}\frac{1}{\cos(10t)}=\frac{1}{\cos(0)} = 1$. Then $\lim_{t
ightarrow0}\frac{\sin(10t)}{10t}\times\frac{10t}{2t}\times\frac{2t}{\sin(2t)}\times\frac{1}{\cos(10t)}=1\times\frac{10}{2}\times1\times1 = 5$.

Answer:

1. $\frac{7}{8}$
2. $5$