given the piecewise function $f(x)=\\begin{cases}1 + 2x+x^{2}&x\\leq2\\\\3&x = 2\\\\\\frac{3x^{2}+6}{x}&2\\lt…

given the piecewise function $f(x)=\\begin{cases}1 + 2x+x^{2}&x\\leq2\\\\3&x = 2\\\\\\frac{3x^{2}+6}{x}&2\\lt x\\lt4\\\\\\frac{1}{x}&x\\geq4\\end{cases}$\ndetermine the following limits:\n(a) $\\lim_{x\\to2^{-}}f(x)$\n(b) $\\lim_{x\\to2^{+}}f(x)$\n(c) $\\lim_{x\\to2}f(x)$\n(d) $\\lim_{x\\to4^{-}}f(x)$\n(e) $\\lim_{x\\to4^{+}}f(x)$\n(f) $\\lim_{x\\to4}f(x)$

given the piecewise function $f(x)=\\begin{cases}1 + 2x+x^{2}&x\\leq2\\\\3&x = 2\\\\\\frac{3x^{2}+6}{x}&2\\lt x\\lt4\\\\\\frac{1}{x}&x\\geq4\\end{cases}$\ndetermine the following limits:\n(a) $\\lim_{x\\to2^{-}}f(x)$\n(b) $\\lim_{x\\to2^{+}}f(x)$\n(c) $\\lim_{x\\to2}f(x)$\n(d) $\\lim_{x\\to4^{-}}f(x)$\n(e) $\\lim_{x\\to4^{+}}f(x)$\n(f) $\\lim_{x\\to4}f(x)$

Answer

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}\neq\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