Facts: \[\lim_{x\to 0} \frac{\sin x}{x}=1\] \[\lim_{x\to 0} \frac{\cos x-1}{x}=0\]
Example 1.
Compute the limit
\[\lim_{x\to 0}\frac{\sin(2x)}{x}.\]
Sol:
\[\lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{\sin(2x)}{2x}\cdot 2=\left(\lim_{x\to 0}\frac{\sin(2x)}{2x}\right) \cdot 2=1\cdot 2 = 2\]
Ex. Find the limit \(\lim\limits_{x\to 0}\frac{\sin(3x)}{5x}\).
Answer: \(\frac{3}{5}\)
Fact: \[\lim\limits_{x\to 0}\frac{\sin(ax)}{bx}=\dfrac{a}{b}\quad (\text{here}\; b\neq 0)\]
Example 2.
Find the derivative of \(f(x)=x^2\cos x\).
Sol: Product rule says that
\[f'(x)=(x^2)'\cos x + x^2(\cos x)'=(2x)\cos x + x^2(-\sin x)=2x\cos x-x^2\sin x\]
Example 3.
Use the fact that \(\frac{d}{dx}(\sin x)=\cos x\) to explain why \(\lim\limits_{x\to 0}\frac{\sin x}{x}=1.\)
Sol:
Let \(f(x)=\sin x\). Then we know that \(f'(x)=\cos x\).
Note that \(f(0)=\sin 0 = 0\) and thus
\[\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\sin x-\sin 0}{x-0}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)=\cos 0 = 1.\]
The third limit above is the definition of the derivative \(f'(0)\).