Question

a. graph f(x)={1 - x^2, x≠ - 1; 2, x = - 1}. b. find lim┬(x→ - 1^ - )⁡f(x) and lim┬(x→ - 1^ + )⁡f(x). c. does lim┬(x→ - 1)⁡f(x) exist? if so, what is it? if not, why not? a. choose the correct graph below.

Explanation:

Step1: Analyze the function for $x

eq - 1$
The function $f(x)=1 - x^{2}$ for $x
eq - 1$. This is a parabola opening downwards with vertex at $(0,1)$.

Step2: Consider the value at $x=-1$

The function has a value of $2$ at $x = - 1$. So there is a point - like discontinuity at $x=-1$.

Step3: Find the left - hand limit

For $\lim_{x
ightarrow - 1^{-}}f(x)$, we use $f(x)=1 - x^{2}$. Substitute values approaching $-1$ from the left into $y = 1 - x^{2}$. $\lim_{x
ightarrow - 1^{-}}(1 - x^{2})=1-(-1)^{2}=0$.

Step4: Find the right - hand limit

For $\lim_{x
ightarrow - 1^{+}}f(x)$, we use $f(x)=1 - x^{2}$. Substitute values approaching $-1$ from the right into $y = 1 - x^{2}$. $\lim_{x
ightarrow - 1^{+}}(1 - x^{2})=1-(-1)^{2}=0$.

Step5: Determine the overall limit

Since $\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{+}}f(x)=0$, but $f(-1)=2$, $\lim_{x
ightarrow - 1}f(x)=0$ and the function is discontinuous at $x=-1$.

Step6: Match with the graph

The graph should show the parabola $y = 1 - x^{2}$ with a hole at $x=-1$ and a point at $(-1,2)$.

Answer:

a. The graph should show a parabola $y = 1 - x^{2}$ with a hole at $x=-1$ and a point $(-1,2)$.
b. $\lim_{x
ightarrow - 1^{-}}f(x)=0$, $\lim_{x
ightarrow - 1^{+}}f(x)=0$
c. $\lim_{x
ightarrow - 1}f(x)=0$ exists because $\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{+}}f(x)=0$, but the function is discontinuous at $x = - 1$ since $f(-1)=2
eq0$.