Question

1. use $lim_{u \to 0} \frac{sin u}{u}=1$ to evaluate (i) $lim_{x \to 0} \frac{sin(2x)}{x}$, (ii) $lim_{x \to 0} \frac{sin(x^{2})}{x}$, (iii) $lim_{x \to 0} \frac{\tan(4x)}{x}$.

Explanation:

Step1: Evaluate $\lim_{x

ightarrow0}\frac{\sin(2x)}{x}$
Let $u = 2x$. As $x
ightarrow0$, $u
ightarrow0$. Then $\lim_{x
ightarrow0}\frac{\sin(2x)}{x}=\lim_{x
ightarrow0}\frac{\sin(2x)}{x}\cdot\frac{2}{2}=2\lim_{x
ightarrow0}\frac{\sin(2x)}{2x}=2\times1 = 2$.

Step2: Evaluate $\lim_{x

ightarrow0}\frac{\sin(x^{2})}{x}$
Rewrite it as $\lim_{x
ightarrow0}\frac{\sin(x^{2})}{x}=\lim_{x
ightarrow0}x\cdot\frac{\sin(x^{2})}{x^{2}}$. Let $u = x^{2}$, as $x
ightarrow0$, $u
ightarrow0$. So $\lim_{x
ightarrow0}x\cdot\frac{\sin(x^{2})}{x^{2}}=\lim_{x
ightarrow0}x\cdot\lim_{u
ightarrow0}\frac{\sin(u)}{u}=0\times1 = 0$.

Step3: Evaluate $\lim_{x

ightarrow0}\frac{\tan(4x)}{x}$
Since $\tan(4x)=\frac{\sin(4x)}{\cos(4x)}$, then $\lim_{x
ightarrow0}\frac{\tan(4x)}{x}=\lim_{x
ightarrow0}\frac{\sin(4x)}{x\cos(4x)}=\lim_{x
ightarrow0}\frac{\sin(4x)}{x}\cdot\frac{1}{\cos(4x)}$. Let $u = 4x$, as $x
ightarrow0$, $u
ightarrow0$. So $\lim_{x
ightarrow0}\frac{\sin(4x)}{x}\cdot\frac{1}{\cos(4x)} = 4\lim_{x
ightarrow0}\frac{\sin(4x)}{4x}\cdot\frac{1}{\cos(4x)}=4\times1\times\frac{1}{1}=4$.

Answer:

(i) $2$
(ii) $0$
(iii) $4$