consider the function\nf(x)=\begin{cases}x^{2}-4, & x < 1,\\\frac{x^{2}+alpha x+\beta}{x - 2}, & 1leq x <…

Explanation:

Step1: Definition of continuity

A function $f(x)$ is continuous at $x = a$ if $\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)$. That is, the left - hand limit, the right - hand limit and the value of the function at $x = a$ are equal.

Step2: Continuity at $x = 1$

First, find $\lim_{x\rightarrow1^{-}}f(x)$ and $\lim_{x\rightarrow1^{+}}f(x)$. $\lim_{x\rightarrow1^{-}}f(x)=1^{2}-4=- 3$. $\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1}\frac{x^{2}+\alpha x+\beta}{x - 2}$. Since the limit exists and the denominator approaches $1 - 2=-1$, then $1^{2}+\alpha\times1+\beta = 0$, so $\alpha+\beta=-1$. And $\lim_{x\rightarrow1^{+}}f(x)=\frac{1+\alpha+\beta}{1 - 2}=3$.

Step3: Continuity at $x = 2$

Find $\lim_{x\rightarrow2^{-}}f(x)$ and $\lim_{x\rightarrow2^{+}}f(x)$. $\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2}\frac{x^{2}+\alpha x+\beta}{x - 2}$. Since $\alpha+\beta=-1$, then $x^{2}+\alpha x+\beta=x^{2}+\alpha x-(1 + \alpha)=(x - 1)(x+(1 + \alpha))$. So $\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2}\frac{(x - 1)(x+(1+\alpha))}{x - 2}$. For the limit to exist, when $x = 2$, the numerator must be $0$, so $2+(1+\alpha)=0$, then $\alpha=-3$. Since $\alpha=-3$, then $\beta = 2$. $\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2}\frac{x^{2}-3x + 2}{x - 2}=\lim_{x\rightarrow2}\frac{(x - 1)(x - 2)}{x - 2}=1$. $\lim_{x\rightarrow2^{+}}f(x)=\gamma e^{2 - 2}=\gamma$. So $\gamma = 1$.

Answer:

(a) A function $f(x)$ is continuous at $x = a$ if $\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)$. (b) $\alpha=-3$, $\beta = 2$, $\gamma = 1$