Question

1. evaluate the limits. write does not exist where appropriate.

a) $lim_{x \to 2}\frac{7x+sqrt3{2x + 4}}{x^{2}+4}$
b) $lim_{x \to 2}\frac{x^{2}-7x + 10}{x^{3}+3x^{2}-10x}$
c) $lim_{x \to 5}\frac{sqrt{x + 4}-3}{x^{2}-25}$

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 \(\lim_{x
ightarrow2}\frac{x^{2}-7x + 10}{x^{3}+3x^{2}-10x}=\lim_{x
ightarrow2}\frac{(x - 2)(x - 5)}{x(x - 2)(x + 5)}\). Cancel out the common - factor \((x - 2)\) (since \(x
eq2\) when taking the limit), we get \(\lim_{x
ightarrow2}\frac{x - 5}{x(x + 5)}\). 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 \(\lim_{x
ightarrow5}\frac{\sqrt{x + 4}-3}{x^{2}-25}=\lim_{x
ightarrow5}\frac{x - 5}{(x - 5)(x + 5)(\sqrt{x + 4}+3)}\). Cancel out the common - factor \((x - 5)\) (since \(x
eq5\) when taking the limit), we get \(\lim_{x
ightarrow5}\frac{1}{(x + 5)(\sqrt{x + 4}+3)}\). 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) \(\frac{1}{60}\)