Question-16

Evaluate:

\[ \begin{equation*} \lim\limits _{x\rightarrow -3} \ \ \frac{\sqrt{2x+22} -4}{x+3} \end{equation*} \]

Answer

\(0.25\)

Solution

We can express the Taylor series expansion for \(\displaystyle \sqrt{2x+22}\) around \(\displaystyle x=-3\) as: \[ \begin{equation*} 4+\frac{1}{4}( x+3) +( x+3)^{2} p( x) \end{equation*} \]

where \(\displaystyle p( x)\) is some polynomial of \(\displaystyle x\).

\[ \begin{equation*} \begin{aligned} \lim\limits _{x\rightarrow -3} \ \ \frac{\sqrt{2x+22} -4}{x+3} & =\lim\limits _{x\rightarrow -3} \ \ \frac{\frac{1}{4}( x+3) +( x+3)^{2} p( x)}{x+3}\\ & \\ & =\frac{1}{4} \end{aligned} \end{equation*} \]