Question

1. determine the values of the following limits.

(a) $lim_{x
ightarrow - 1}\frac{x^{2}+3x + 2}{x^{2}+4x + 3}$
(b) $lim_{x
ightarrow0}\frac{3-sqrt{2x + 9}}{x}$
(c) $lim_{x
ightarrow2}\frac{(\frac{1}{x - 1}-\frac{3}{x + 1})}{x - 2}$

Explanation:

Step1: Factor the numerator and denominator for (a)

For $\lim_{x
ightarrow - 1}\frac{x^{2}+3x + 2}{x^{2}+4x + 3}$, factor $x^{2}+3x + 2=(x + 1)(x+2)$ and $x^{2}+4x + 3=(x + 1)(x + 3)$. Then the limit becomes $\lim_{x
ightarrow - 1}\frac{(x + 1)(x + 2)}{(x + 1)(x + 3)}$. Cancel out the common factor $(x + 1)$ (since $x
eq - 1$ when taking the limit), we get $\lim_{x
ightarrow - 1}\frac{x + 2}{x + 3}$.

Step2: Evaluate the limit for (a)

Substitute $x=-1$ into $\frac{x + 2}{x + 3}$, we have $\frac{-1+2}{-1 + 3}=\frac{1}{2}$.

Step3: Rationalize the numerator for (b)

For $\lim_{x
ightarrow0}\frac{3-\sqrt{2x + 9}}{x}$, multiply the numerator and denominator by the conjugate of the numerator $3+\sqrt{2x + 9}$. We get $\lim_{x
ightarrow0}\frac{(3-\sqrt{2x + 9})(3+\sqrt{2x + 9})}{x(3+\sqrt{2x + 9})}$. Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator is $9-(2x + 9)=-2x$. So the limit becomes $\lim_{x
ightarrow0}\frac{-2x}{x(3+\sqrt{2x + 9})}$. Cancel out the common factor $x$ (since $x
eq0$ when taking the limit), we have $\lim_{x
ightarrow0}\frac{-2}{3+\sqrt{2x + 9}}$.

Step4: Evaluate the limit for (b)

Substitute $x = 0$ into $\frac{-2}{3+\sqrt{2x + 9}}$, we get $\frac{-2}{3+\sqrt{9}}=\frac{-2}{3 + 3}=-\frac{1}{3}$.

Step5: Combine the fractions in the numerator for (c)

For $\lim_{x
ightarrow2}\frac{\frac{1}{x - 1}-\frac{3}{x + 1}}{x - 2}$, first combine the fractions in the numerator: $\frac{1}{x - 1}-\frac{3}{x + 1}=\frac{(x + 1)-3(x - 1)}{(x - 1)(x + 1)}=\frac{x + 1-3x + 3}{(x - 1)(x + 1)}=\frac{-2x+4}{(x - 1)(x + 1)}$. So the original limit is $\lim_{x
ightarrow2}\frac{\frac{-2x + 4}{(x - 1)(x + 1)}}{x - 2}=\lim_{x
ightarrow2}\frac{-2x + 4}{(x - 1)(x + 1)(x - 2)}$. Factor out $-2$ from the numerator: $\frac{-2(x - 2)}{(x - 1)(x + 1)(x - 2)}$. Cancel out the common factor $(x - 2)$ (since $x
eq2$ when taking the limit), we get $\lim_{x
ightarrow2}\frac{-2}{(x - 1)(x + 1)}$.

Step6: Evaluate the limit for (c)

Substitute $x = 2$ into $\frac{-2}{(x - 1)(x + 1)}$, we have $\frac{-2}{(2 - 1)(2+1)}=-\frac{2}{3}$.

Answer:

(a) $\frac{1}{2}$
(b) $-\frac{1}{3}$
(c) $-\frac{2}{3}$