2. evaluate the limits. write does not exist where appropriate.\na) $lim_{x \to 2}\frac{7x+sqrt3{2x +…

Explanation:

Step1: Substitute (x = 2) into part a

Substitute (x = 2) into (\frac{7x+\sqrt[3]{2x + 4}}{x^{2}+4}). [ \begin{align*} \frac{7\times2+\sqrt[3]{2\times2 + 4}}{2^{2}+4}&=\frac{14+\sqrt[3]{4 + 4}}{4 + 4}\ &=\frac{14+\sqrt[3]{8}}{8}\ &=\frac{14 + 2}{8}\ &=\frac{16}{8}\ &=2 \end{align*} ]

Step2: Factor the numerator and denominator in part b

Factor (x^{2}-7x + 10=(x - 2)(x - 5)) and (x^{3}+3x^{2}-10x=x(x^{2}+3x - 10)=x(x - 2)(x+5)). Then (\lim_{x\rightarrow2}\frac{x^{2}-7x + 10}{x^{3}+3x^{2}-10x}=\lim_{x\rightarrow2}\frac{(x - 2)(x - 5)}{x(x - 2)(x + 5)}). Cancel out the common - factor ((x - 2)) (since (x\neq2) when taking the limit), we get (\lim_{x\rightarrow2}\frac{x - 5}{x(x + 5)}). Now substitute (x = 2): (\frac{2-5}{2\times(2 + 5)}=\frac{-3}{14}).

Step3: Rationalize the numerator in part c

Multiply the numerator and denominator of (\frac{\sqrt{x + 4}-3}{x^{2}-25}) by (\sqrt{x + 4}+3). The numerator becomes ((\sqrt{x + 4}-3)(\sqrt{x + 4}+3)=(x + 4)-9=x - 5), and the denominator is ((x^{2}-25)(\sqrt{x + 4}+3)=(x - 5)(x + 5)(\sqrt{x + 4}+3)). Then (\lim_{x\rightarrow5}\frac{\sqrt{x + 4}-3}{x^{2}-25}=\lim_{x\rightarrow5}\frac{x - 5}{(x - 5)(x + 5)(\sqrt{x + 4}+3)}). Cancel out the common - factor ((x - 5)) (since (x\neq5) when taking the limit), we get (\lim_{x\rightarrow5}\frac{1}{(x + 5)(\sqrt{x + 4}+3)}). Substitute (x = 5): (\frac{1}{(5 + 5)(\sqrt{5+4}+3)}=\frac{1}{10\times(3 + 3)}=\frac{1}{60}).

Answer:

a) (2) b) (-\frac{3}{14}) c) (\frac{1}{60})