Chapter 6: Problem 4

Draw a graph of the function $$ f(x)= \begin{cases}2 x+1 & x<3 \\ 5 & x=3 \\ 6 & x>3\end{cases} $$ Find (a) $\lim _{x \rightarrow 0^{+}} f(x)$ (b) $\lim _{x \rightarrow 0}-f(x)$ (c) $\lim _{x \rightarrow 0} f(x)$ (d) $\lim _{x \rightarrow 3^{+}} f(x)$ (e) $\lim _{x \rightarrow 3^{-}} f(x)$ (f) $\lim _{x \rightarrow 3} f(x)$

Short Answer

Expert verified

Answer: The limits of the function are as follows: 1. $\lim_{x\rightarrow 0} f(x) = 1$ 2. $\lim_{x\rightarrow 0^+} f(x) = 1$ 3. $\lim_{x\rightarrow 0^-} f(x) = 1$ 4. $\lim_{x\rightarrow 3} f(x)$ does not exist. 5. $\lim_{x\rightarrow 3^+} f(x) = 6$ 6. $\lim_{x\rightarrow 3^-} f(x) = 7$

Step by step solution

Draw the graph of the function

First, we will need to draw a graph of the given function. Notice that the function has three different cases: 1. When $x<3$, the function is $f(x) = 2x + 1$ 2. When $x=3$, the function is $f(x) = 5$ 3. When $x>3$, the function is $f(x) = 6$ In this case, to draw the graph on a piece of paper or using a graphing software, draw three different line segments corresponding to those cases above.

Find the limit when x approaches 0

(a) $\lim _{x \rightarrow 0^{+}} f(x)$: Since $0 < 3$, we use the first case $f(x) = 2x + 1$. So the limit becomes: $$\lim _{x \rightarrow 0^{+}} (2x + 1)$$ As $x$ approaches 0, the expression becomes: $$2(0) + 1 = 1$$ So the limit is 1. (b) $\lim _{x \rightarrow 0^{-}} f(x)$: Again, since $0 < 3$, we use the first case $f(x) = 2x + 1$. So the limit becomes: $$\lim _{x \rightarrow 0^{-}} (2x + 1)$$ As $x$ approaches 0, the expression becomes: $$2(0) + 1 = 1$$ So the limit is 1. (c) $\lim _{x \rightarrow 0} f(x)$: Since both the limits from the right and left are equal to 1, the limit at 0 exists and is equal to 1.

Find the limit when x approaches 3

(d) $\lim _{x \rightarrow 3^{+}} f(x)$: For $x > 3$, we use the third case $f(x) = 6$. So the limit becomes: $$\lim _{x \rightarrow 3^{+}} 6$$ Since the function is a constant in this interval, the limit is equal to the constant, which is 6. (e) $\lim _{x \rightarrow 3^{-}} f(x)$: For $x < 3$, we use the first case $f(x) = 2x + 1$. So the limit becomes: $$\lim _{x \rightarrow 3^{-}} (2x + 1)$$ As $x$ approaches 3, the expression becomes: $$2(3) + 1 = 7$$ So the limit is 7. (f) $\lim _{x \rightarrow 3} f(x)$: Since the limits from the right and left are not equal (6 and 7), the limit at 3 does not exist.

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Short Answer

Step by step solution

Draw the graph of the function

Find the limit when x approaches 0

Find the limit when x approaches 3

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