Question

(8) 4. evaluate the limits
a) $lim_{x
ightarrow4}\frac{x^{2}+1+sqrt{2x + 1}}{x^{2}-5x + 6}$
b) $lim_{x
ightarrow2}\frac{x^{3}-4x}{x^{2}+x - 6}$
c) $lim_{x
ightarrow2}\frac{sqrt{x + 7}-3}{x^{2}+x - 6}$
d) $lim_{x
ightarrow3}\frac{1}{x^{2}-3x}-\frac{2}{x^{2}-9}$

Explanation:

Step1: Evaluate numerator and denominator for a)

For $\lim_{x
ightarrow4}\frac{x^{2}+1 + \sqrt{2x + 1}}{x^{2}-5x + 6}$, when $x = 4$:
The numerator is $4^{2}+1+\sqrt{2\times4 + 1}=16 + 1+\sqrt{9}=16+1 + 3=20$.
The denominator is $4^{2}-5\times4 + 6=16-20 + 6=2$.

Step2: Calculate limit for a)

$\lim_{x
ightarrow4}\frac{x^{2}+1+\sqrt{2x + 1}}{x^{2}-5x + 6}=\frac{20}{2}=10$.

Step3: Factor expressions for b)

For $\lim_{x
ightarrow2}\frac{x^{3}-4x}{x^{2}+x - 6}$, factor $x^{3}-4x=x(x^{2}-4)=x(x - 2)(x + 2)$ and $x^{2}+x - 6=(x - 2)(x+3)$.

Step4: Simplify and calculate limit for b)

$\lim_{x
ightarrow2}\frac{x(x - 2)(x + 2)}{(x - 2)(x + 3)}=\lim_{x
ightarrow2}\frac{x(x + 2)}{x + 3}=\frac{2\times(2 + 2)}{2+3}=\frac{8}{5}$.

Step5: Rationalize numerator for c)

For $\lim_{x
ightarrow2}\frac{\sqrt{x + 7}-3}{x^{2}+x - 6}$, multiply numerator and denominator by $\sqrt{x + 7}+3$.
The numerator becomes $(x + 7)-9=x - 2$.
The denominator is $(x - 2)(x + 3)$.

Step6: Simplify and calculate limit for c)

$\lim_{x
ightarrow2}\frac{x - 2}{(x - 2)(x + 3)(\sqrt{x + 7}+3)}=\lim_{x
ightarrow2}\frac{1}{(x + 3)(\sqrt{x + 7}+3)}=\frac{1}{(2 + 3)(\sqrt{2+7}+3)}=\frac{1}{5\times(3 + 3)}=\frac{1}{30}$.

Step7: Find common - denominator for d)

For $\lim_{x
ightarrow3}(\frac{1}{x^{2}-3x}-\frac{2}{x^{2}-9})$, factor $x^{2}-3x=x(x - 3)$ and $x^{2}-9=(x - 3)(x + 3)$.
The common denominator is $x(x - 3)(x + 3)$.
$\frac{1}{x^{2}-3x}-\frac{2}{x^{2}-9}=\frac{x + 3-2x}{x(x - 3)(x + 3)}=\frac{3 - x}{x(x - 3)(x + 3)}=-\frac{1}{x(x + 3)}$.

Step8: Calculate limit for d)

$\lim_{x
ightarrow3}-\frac{1}{x(x + 3)}=-\frac{1}{3\times(3 + 3)}=-\frac{1}{18}$.

Answer:

a) $10$
b) $\frac{8}{5}$
c) $\frac{1}{30}$
d) $-\frac{1}{18}$