# Recitation 3

Exercise: Prove that $\lim_{x\rightarrow -3}(1-4x)=13$

Proof. For every $\epsilon > 0$, we pick $\delta = \frac{\epsilon}{4}$. Whenever $|x-(-3)|<\delta$, we will have $|(1-4x)-13|=|-4x-12|=4|x+3|< 4\delta = \epsilon$. Thus $\lim_{x\rightarrow -3}(1-4x)=13$.

Exercise: Calculate the following:

1. $\lim_{x\rightarrow 5}\frac{x^2-6x+5}{x-5}$.
2. $\lim_{h\rightarrow 0}\frac{(x+h)^3-x^3}{h}$.
3. $\lim_{x\rightarrow 0}\frac{\sin{3x}}{x}$.
4. $\lim_{x\rightarrow 0}\frac{\sin{3x}}{5x^3-4x}$.

Proof.

1. When $x\neq 5$, $\frac{x^2-6x+5}{x-5}=x-1$. Thus $\lim_{x\rightarrow 5}\frac{x^2-6x+5}{x-5}=\lim_{x\rightarrow 5}{x-1}=4$.
2. Expand the numerator first, $(x+h)^3-x^3=x^3+3x^2h+3xh^2+h^3-x^3=3x^2h+3xh^2+h^2$. Hence $\lim_{h\rightarrow 0}\frac{(x+h)^3-x^3}{h}=\lim_{h\rightarrow 0}{3x^2+3xh+h^2}=3x^2$
3. As $\lim_{x\rightarrow 0}\frac{\sin{3x}}{3x}=1$, $\lim_{x\rightarrow 0}\frac{\sin{3x}}{x}=3$.
4. As $\frac{\sin{3x}}{5x^3-4x}=\frac{\sin 3x}{3x}\frac{3}{5x^2-4}$, $\lim_{x\rightarrow 0}\frac{\sin{3x}}{5x^3-4x}=\lim_{x\rightarrow 0}\frac{\sin{3x}}{3x}\lim_{x\rightarrow 0}\frac{3}{5x^2-4}=-\frac{3}{4}$