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 $\lim_{x \to 0}\frac{\sin(2x)}{x}$

Let $u = 2x$. As $x\to0$, $u\to0$. Then $\lim_{x \to 0}\frac{\sin(2x)}{x}=\lim_{x \to 0}\frac{\sin(2x)}{x}\cdot\frac{2}{2}=2\lim_{u \to 0}\frac{\sin(u)}{u}$. Since $\lim_{u \to 0}\frac{\sin(u)}{u}=1$, we have $2\times1 = 2$.

Step2: Evaluate $\lim_{x \to 0}\frac{\sin(x^{2})}{x}$

We know that $\lim_{x \to 0}\frac{\sin(x^{2})}{x}=\lim_{x \to 0}\frac{\sin(x^{2})}{x^{2}}\cdot x$. Let $u = x^{2}$. As $x\to0$, $u\to0$. So $\lim_{x \to 0}\frac{\sin(x^{2})}{x^{2}}\cdot x=\lim_{u \to 0}\frac{\sin(u)}{u}\cdot\lim_{x \to 0}x$. Since $\lim_{u \to 0}\frac{\sin(u)}{u}=1$ and $\lim_{x \to 0}x = 0$, the result is $1\times0=0$.

Step3: Evaluate $\lim_{x \to 0}\frac{\tan(4x)}{x}$

Since $\tan(4x)=\frac{\sin(4x)}{\cos(4x)}$, then $\lim_{x \to 0}\frac{\tan(4x)}{x}=\lim_{x \to 0}\frac{\sin(4x)}{x\cos(4x)}=\lim_{x \to 0}\frac{\sin(4x)}{4x}\cdot\frac{4}{\cos(4x)}$. Let $u = 4x$. As $x\to0$, $u\to0$. So $\lim_{x \to 0}\frac{\sin(4x)}{4x}\cdot\frac{4}{\cos(4x)}=\lim_{u \to 0}\frac{\sin(u)}{u}\cdot\lim_{x \to 0}\frac{4}{\cos(4x)}$. Since $\lim_{u \to 0}\frac{\sin(u)}{u}=1$ and $\lim_{x \to 0}\cos(4x)=\cos(0) = 1$, the result is $1\times4 = 4$.

Answer:

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