Question

1. find

lim _ { t
ightarrow 9 } \frac { t - 9 } { sqrt { t } - 3 }

1. find

lim _ { x
ightarrow 1 / 2 } \frac { 2 x ^ { 2 } + 3 x - 2 } { 2 x - 1 }

Explanation:

Step1: Rationalize the denominator for the first limit

Multiply numerator and denominator by $\sqrt{t}+3$:

$$\lim_{t ightarrow9}\frac{t - 9}{\sqrt{t}-3}=\lim_{t ightarrow9}\frac{(t - 9)(\sqrt{t}+3)}{(\sqrt{t}-3)(\sqrt{t}+3)}$$

Since $(a - b)(a + b)=a^{2}-b^{2}$, then $(\sqrt{t}-3)(\sqrt{t}+3)=t - 9$. So the expression becomes:

$$\lim_{t ightarrow9}\frac{(t - 9)(\sqrt{t}+3)}{t - 9}=\lim_{t ightarrow9}(\sqrt{t}+3)$$

Step2: Evaluate the simplified limit

Substitute $t = 9$ into $\sqrt{t}+3$:
$\sqrt{9}+3=3 + 3=6$

Step3: Factor the numerator for the second limit

Factor $2x^{2}+3x - 2=(2x - 1)(x + 2)$
So $\lim_{x
ightarrow\frac{1}{2}}\frac{2x^{2}+3x - 2}{2x - 1}=\lim_{x
ightarrow\frac{1}{2}}\frac{(2x - 1)(x + 2)}{2x - 1}$
Cancel out the common factor $(2x - 1)$ (for $x
eq\frac{1}{2}$ as we are taking the limit):

$$\lim_{x ightarrow\frac{1}{2}}(x + 2)$$

Step4: Evaluate the simplified limit

Substitute $x=\frac{1}{2}$ into $x + 2$:
$\frac{1}{2}+2=\frac{1 + 4}{2}=\frac{5}{2}$

Answer:

For $\lim_{t
ightarrow9}\frac{t - 9}{\sqrt{t}-3}$, the answer is $6$.
For $\lim_{x
ightarrow\frac{1}{2}}\frac{2x^{2}+3x - 2}{2x - 1}$, the answer is $\frac{5}{2}$.