8. find\n lim _ { t \rightarrow 9 } \frac { t - 9 } { sqrt { t } - 3 } \n9. find\n lim _ { x \rightarrow 1 /…

Explanation:

Step1: Rationalize the denominator for the first limit

Multiply numerator and denominator by $\sqrt{t}+3$: $$\lim_{t\rightarrow9}\frac{t - 9}{\sqrt{t}-3}=\lim_{t\rightarrow9}\frac{(t - 9)(\sqrt{t}+3)}{(\sqrt{t}-3)(\sqrt{t}+3)}$$ Since $(a - b)(a + b)=a^{2}-b^{2}$, then $(\sqrt{t}-3)(\sqrt{t}+3)=t - 9$. So the expression becomes: $$\lim_{t\rightarrow9}\frac{(t - 9)(\sqrt{t}+3)}{t - 9}=\lim_{t\rightarrow9}(\sqrt{t}+3)$$

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

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 $\lim_{t\rightarrow9}\frac{t - 9}{\sqrt{t}-3}$, the answer is $6$. For $\lim_{x\rightarrow\frac{1}{2}}\frac{2x^{2}+3x - 2}{2x - 1}$, the answer is $\frac{5}{2}$.