Question

Evaluate the following limits.
$$\displaystyle\lim_{x\rightarrow 3}\left(\dfrac{1}{x-3}-\dfrac{2}{x^2-4x+3}\right)$$.

Answer 1

Correct option is A. 0.5

$$\displaystyle \lim_{x\rightarrow 3}{\left(\dfrac{1}{(x-3)}-\dfrac{2}{x^2-4x+3}\right)}$$

$$\displaystyle \lim_{x\rightarrow 3}{\left(\dfrac{1}{x-3}-\dfrac{2}{(x-1)(x-3)}\right)}$$

$$\displaystyle \lim_{x\rightarrow 3}{\left(\dfrac{(x-1)-2}{(x-1)(x-3)}\right)}$$

$$\displaystyle \lim_{x\rightarrow 3}{\dfrac{(x-3)}{(x-1)(x-3)}}$$

$$\displaystyle \lim_{x\rightarrow 3}{\dfrac{1}{x-1}}=\dfrac{1}{2}=0.5$$