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}