question 7 (1 point)\nconsider the function\nf(x)=\begin{cases}x^{2}+1&x < 1\\2&x = 1\\\frac{x}{2}+c&x>1end{c…

Explanation:

Step1: Calculate left - hand limit

For $x<1$, $f(x)=x^{2}+1$. So, $\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{-}}(x^{2}+1)=1^{2}+1 = 2$.

Step2: Calculate right - hand limit

For $x > 1$, $f(x)=\frac{x}{2}+c$. So, $\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1^{+}}(\frac{x}{2}+c)=\frac{1}{2}+c$.

Step3: Analyze continuity condition

A function $y = f(x)$ is continuous at $x = a$ if $\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)$. Here $f(1) = 2$. For continuity, $\frac{1}{2}+c=2$, which gives $c=\frac{3}{2}$.

Step4: Analyze limit when $c=\frac{1}{2}$

When $c=\frac{1}{2}$, $\lim_{x\rightarrow1^{+}}f(x)=\frac{1}{2}+\frac{1}{2}=1$, and $\lim_{x\rightarrow1^{-}}f(x)=2$. So, $\lim_{x\rightarrow1}f(x)$ does not exist when $c = \frac{1}{2}$.

Step5: Analyze infinite discontinuity

An infinite discontinuity occurs when either $\lim_{x\rightarrow a^{-}}f(x)=\pm\infty$ or $\lim_{x\rightarrow a^{+}}f(x)=\pm\infty$. Since $\lim_{x\rightarrow1^{-}}f(x)=2$ and $\lim_{x\rightarrow1^{+}}f(x)=\frac{1}{2}+c$ (a finite - valued function for any real $c$), there are no values of $c$ for which $f(x)$ has an infinite discontinuity at $x = 1$.

Step6: Analyze removable discontinuity

A removable discontinuity occurs when $\lim_{x\rightarrow a}f(x)$ exists but $f(a)$ is either not defined or not equal to the limit. When $c = 3$, $\lim_{x\rightarrow1^{+}}f(x)=\frac{1}{2}+3=\frac{7}{2}$, $\lim_{x\rightarrow1^{-}}f(x)=2$. Since the left - hand and right - hand limits are not equal, it is not a removable discontinuity.

Step7: Analyze jump discontinuity

A jump discontinuity occurs when $\lim_{x\rightarrow a^{-}}f(x)\neq\lim_{x\rightarrow a^{+}}f(x)$. When $c = 1$, $\lim_{x\rightarrow1^{+}}f(x)=\frac{1}{2}+1=\frac{3}{2}$, $\lim_{x\rightarrow1^{-}}f(x)=2$. So, there is a jump discontinuity.

Answer:

B. When $c=\frac{1}{2},\lim_{x\rightarrow1}f(x)=2$.