Question

evaluate or determine that the limit does not exist for each of the limits (a) $lim_{x
ightarrow d -}f(x)$, (b) $lim_{x
ightarrow d +}f(x)$, and (c) $lim_{x
ightarrow d}f(x)$ for the given function $f$ and number $d$.

1. $f(x)=\begin{cases}x^{2}-2, &\text{for }x < 0\\1, &\text{for }xgeq0end{cases}$, $d = - 4$
2. $f(x)=\begin{cases}-4x - 3, &\text{for }x < 1\\1, &\text{for }x = 1\\-2x - 3, &\text{for }x>1end{cases}$, $d = 1$
3. $f(x)=\begin{cases}-5x - 1, &\text{for }xleq1\\-7x + 1, &\text{for }x>1end{cases}$, $d = 1$
4. $f(x)=\begin{cases}\frac{1}{x - 3}, &\text{for }x>3\\x^{2}-4x, &\text{for }xleq3end{cases}$, $d = 3$

Explanation:

Step1: Find left - hand limit for problem 47

For $x\to - 4^-$, since $x=-4<0$, we use $f(x)=x^{2}-2$. Substitute $x = - 4$ into $x^{2}-2$.
$(-4)^{2}-2=16 - 2=14$

Step2: Find right - hand limit for problem 47

For $x\to - 4^+$, since $x=-4<0$, we use $f(x)=x^{2}-2$. Substitute $x=-4$ into $x^{2}-2$.
$(-4)^{2}-2 = 14$

Step3: Find overall limit for problem 47

Since $\lim_{x\to - 4^-}f(x)=\lim_{x\to - 4^+}f(x)=14$, then $\lim_{x\to - 4}f(x)=14$

Step4: Find left - hand limit for problem 48

For $x\to1^-$, since $x < 1$, we use $f(x)=-4x - 3$. Substitute $x = 1$ into $-4x - 3$.
$-4\times1-3=-4 - 3=-7$

Step5: Find right - hand limit for problem 48

For $x\to1^+$, since $x>1$, we use $f(x)=-2x - 3$. Substitute $x = 1$ into $-2x - 3$.
$-2\times1-3=-2 - 3=-5$

Step6: Find overall limit for problem 48

Since $\lim_{x\to1^-}f(x)=-7$ and $\lim_{x\to1^+}f(x)=-5$, $\lim_{x\to1}f(x)$ does not exist.

Step7: Find left - hand limit for problem 49

For $x\to1^-$, since $x\leq1$, we use $f(x)=-5x - 1$. Substitute $x = 1$ into $-5x - 1$.
$-5\times1-1=-5 - 1=-6$

Step8: Find right - hand limit for problem 49

For $x\to1^+$, since $x>1$, we use $f(x)=-7x + 1$. Substitute $x = 1$ into $-7x + 1$.
$-7\times1+1=-7 + 1=-6$

Step9: Find overall limit for problem 49

Since $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=-6$, then $\lim_{x\to1}f(x)=-6$

Step10: Find left - hand limit for problem 50

For $x\to3^-$, since $x\leq3$, we use $f(x)=x^{2}-4x$. Substitute $x = 3$ into $x^{2}-4x$.
$3^{2}-4\times3=9 - 12=-3$

Step11: Find right - hand limit for problem 50

For $x\to3^+$, since $x>3$, we use $f(x)=\frac{1}{x - 3}$. As $x\to3^+$, $\lim_{x\to3^+}\frac{1}{x - 3}=\infty$.

Step12: Find overall limit for problem 50

Since $\lim_{x\to3^-}f(x)=-3$ and $\lim_{x\to3^+}f(x)=\infty$, $\lim_{x\to3}f(x)$ does not exist.

Answer:

47:
a. $\lim_{x\to - 4^-}f(x)=14$
b. $\lim_{x\to - 4^+}f(x)=14$
c. $\lim_{x\to - 4}f(x)=14$
48:
a. $\lim_{x\to1^-}f(x)=-7$
b. $\lim_{x\to1^+}f(x)=-5$
c. $\lim_{x\to1}f(x)$ does not exist
49:
a. $\lim_{x\to1^-}f(x)=-6$
b. $\lim_{x\to1^+}f(x)=-6$
c. $\lim_{x\to1}f(x)=-6$
50:
a. $\lim_{x\to3^-}f(x)=-3$
b. $\lim_{x\to3^+}f(x)=\infty$
c. $\lim_{x\to3}f(x)$ does not exist