Question

(a) $lim_{x
ightarrow3}h(x)$ where $h(x)=\begin{cases}\frac{x^{2}-5x + 6}{x^{2}+6x - 27}&\text{if }x < 3\\frac{sqrt{x + 1}-2}{x^{2}-3x}&\text{if }xgeq3end{cases}$

Explanation:

Step1: Find the left - hand limit

We need to find $\lim_{x
ightarrow3^{-}}h(x)$. Since $x < 3$, $h(x)=\frac{x^{2}-5x + 6}{x^{2}+6x - 27}$.
Factor the numerator and denominator: $x^{2}-5x + 6=(x - 2)(x - 3)$ and $x^{2}+6x - 27=(x + 9)(x - 3)$.
So, $\lim_{x
ightarrow3^{-}}\frac{x^{2}-5x + 6}{x^{2}+6x - 27}=\lim_{x
ightarrow3^{-}}\frac{(x - 2)(x - 3)}{(x + 9)(x - 3)}=\lim_{x
ightarrow3^{-}}\frac{x - 2}{x + 9}$.
Substitute $x = 3$ into $\frac{x - 2}{x + 9}$, we get $\frac{3-2}{3 + 9}=\frac{1}{12}$.

Step2: Find the right - hand limit

We need to find $\lim_{x
ightarrow3^{+}}h(x)$. Since $x\geq3$, $h(x)=\frac{\sqrt{x + 1}-2}{x^{2}-3x}$.
Rationalize the numerator: Multiply the numerator and denominator by $\sqrt{x + 1}+2$.
The numerator becomes $(\sqrt{x + 1}-2)(\sqrt{x + 1}+2)=(x + 1)-4=x - 3$.
The denominator is $x(x - 3)$.
So, $\lim_{x
ightarrow3^{+}}\frac{\sqrt{x + 1}-2}{x^{2}-3x}=\lim_{x
ightarrow3^{+}}\frac{x - 3}{x(x - 3)(\sqrt{x + 1}+2)}=\lim_{x
ightarrow3^{+}}\frac{1}{x(\sqrt{x + 1}+2)}$.
Substitute $x = 3$ into $\frac{1}{x(\sqrt{x + 1}+2)}$, we get $\frac{1}{3(\sqrt{3 + 1}+2)}=\frac{1}{3(2 + 2)}=\frac{1}{12}$.

Step3: Determine the limit

Since $\lim_{x
ightarrow3^{-}}h(x)=\lim_{x
ightarrow3^{+}}h(x)=\frac{1}{12}$, then $\lim_{x
ightarrow3}h(x)=\frac{1}{12}$.

Answer:

$\frac{1}{12}$