Question

given the piecewise function $f(x)=\

$$\begin{cases}1 + 2x+x^{2}&x\\leq2\\\\3&x = 2\\\\\\frac{3x^{2}+6}{x}&2\\lt x\\lt4\\\\\\frac{1}{x}&x\\geq4\\end{cases}$$

$
determine the following limits:
(a) $\lim_{x\to2^{-}}f(x)$
(b) $\lim_{x\to2^{+}}f(x)$
(c) $\lim_{x\to2}f(x)$
(d) $\lim_{x\to4^{-}}f(x)$
(e) $\lim_{x\to4^{+}}f(x)$
(f) $\lim_{x\to4}f(x)$

Explanation:

Step1: Left - hand limit as \(x\to2\)

For \(x\to2^{-}\), use \(f(x)=1 + 2x+x^{2}\).
Substitute \(x = 2\) into \(1+2x+x^{2}\):
\(1+2\times2 + 2^{2}=1 + 4+4=9\)

Step2: Right - hand limit as \(x\to2\)

For \(x\to2^{+}\), use \(f(x)=\frac{3x^{2}+6}{x}\).
Substitute \(x = 2\) into \(\frac{3x^{2}+6}{x}\):
\(\frac{3\times2^{2}+6}{2}=\frac{12 + 6}{2}=\frac{18}{2}=9\)

Step3: Limit as \(x\to2\)

Since \(\lim_{x\to2^{-}}f(x)=\lim_{x\to2^{+}}f(x)=9\), then \(\lim_{x\to2}f(x)=9\)

Step4: Left - hand limit as \(x\to4\)

For \(x\to4^{-}\), use \(f(x)=\frac{3x^{2}+6}{x}\).
Substitute \(x = 4\) into \(\frac{3x^{2}+6}{x}\):
\(\frac{3\times4^{2}+6}{4}=\frac{48+6}{4}=\frac{54}{4}=\frac{27}{2}\)

Step5: Right - hand limit as \(x\to4\)

For \(x\to4^{+}\), use \(f(x)=\frac{1}{x}\).
Substitute \(x = 4\) into \(\frac{1}{x}\): \(\frac{1}{4}\)

Step6: Limit as \(x\to4\)

Since \(\lim_{x\to4^{-}}f(x)=\frac{27}{2}\) and \(\lim_{x\to4^{+}}f(x)=\frac{1}{4}\), and \(\frac{27}{2}
eq\frac{1}{4}\), \(\lim_{x\to4}f(x)\) does not exist.

Answer:

(a) \(9\)
(b) \(9\)
(c) \(9\)
(d) \(\frac{27}{2}\)
(e) \(\frac{1}{4}\)
(f) Does not exist