Evaluate the limits.
f(x)=⎧⎩⎨⎪⎪5x2−65x3+65x2+6x<−1−1≤x≤1x>1
Enter DNE if the limit does not exist.
a) limx→−1−f(x) = 

b) limx→−1+f(x) = 

c) limx→−1f(x) = 

d) f(−1) = 

e) limx→1−f(x) = 

f) limx→1+f(x) = 

g) limx→1f(x) = 

h) f(1) = 

We are given a piecewise function:

\[
f(x) = \begin{cases} 
5x^2 - 6 & \text{if } x < -1 \\
5x^3 + 6 & \text{if } -1 \leq x \leq 1 \\
x & \text{if } x > 1 
\end{cases}
\]

### a) \(\lim_{x \to -1^-} f(x)\)
This is the left-hand limit as \(x\) approaches \(-1\). For \(x < -1\), the function is defined as \(f(x) = 5x^2 - 6\). We substitute \(x = -1\) into this expression:

\[
\lim_{x \to -1^-} f(x) = 5(-1)^2 - 6 = 5(1) - 6 = -1
\]

### b) \(\lim_{x \to -1^+} f(x)\)
This is the right-hand limit as \(x\) approaches \(-1\). For \(-1 \leq x \leq 1\), the function is defined as \(f(x) = 5x^3 + 6\). We substitute \(x = -1\) into this expression:

\[
\lim_{x \to -1^+} f(x) = 5(-1)^3 + 6 = 5(-1) + 6 = 1
\]

### c) \(\lim_{x \to -1} f(x)\)
Since the left-hand and right-hand limits are different (\(-1\) and \(1\), respectively), the limit does not exist at \(x = -1\):

\[
\lim_{x \to -1} f(x) = \text{DNE}
\]

### d) \(f(-1)\)
The value of the function at \(x = -1\) is found from the second piece, since \(-1 \leq x \leq 1\) applies here. Thus,

\[
f(-1) = 5(-1)^3 + 6 = 5(-1) + 6 = 1
\]

### e) \(\lim_{x \to 1^-} f(x)\)
This is the left-hand limit as \(x\) approaches \(1\). For \(-1 \leq x \leq 1\), the function is \(f(x) = 5x^3 + 6\). We substitute \(x = 1\) into this expression:

\[
\lim_{x \to 1^-} f(x) = 5(1)^3 + 6 = 5(1) + 6 = 11
\]

### f) \(\lim_{x \to 1^+} f(x)\)
This is the right-hand limit as \(x\) approaches \(1\). For \(x > 1\), the function is \(f(x) = x\). We substitute \(x = 1\):

\[
\lim_{x \to 1^+} f(x) = 1
\]

### g) \(\lim_{x \to 1} f(x)\)
Since the left-hand and right-hand limits are different (\(11\) and \(1\), respectively), the limit does not exist at \(x = 1\):

\[
\lim_{x \to 1} f(x) = \text{DNE}
\]

### h) \(f(1)\)
The value of the function at \(x = 1\) is given by the second piece, where \(-1 \leq x \leq 1\):

\[
f(1) = 5(1)^3 + 6 = 5(1) + 6 = 11
\]

### Final Answers:
a) \(\lim_{x \to -1^-} f(x) = -1\)  
b) \(\lim_{x \to -1^+} f(x) = 1\)  
c) \(\lim_{x \to -1} f(x) = \text{DNE}\)  
d) \(f(-1) = 1\)  
e) \(\lim_{x \to 1^-} f(x) = 11\)  
f) \(\lim_{x \to 1^+} f(x) = 1\)  
g) \(\lim_{x \to 1} f(x) = \text{DNE}\)  
h) \(f(1) = 11\)

Do you need further details or have any questions?

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Here are some related questions to expand your understanding:
1. What are the conditions for a limit to exist at a point?
2. How can a piecewise function have discontinuities at specific points?
3. Can the function still be continuous even if one of its piecewise components is a simple function (e.g., a polynomial)?
4. How would the limits change if the middle piece were a different polynomial?
5. What is the difference between jump discontinuity and removable discontinuity?

**Tip:** When evaluating limits of piecewise functions, always consider the behavior from both sides of the point where the function definition changes.

Evaluate the limits.
f(x) = ⎧⎩⎨⎪⎪5x^2−6 for x < −1, 5x^3+6 for −1 ≤ x ≤ 1, x for x > 1
Enter DNE if the limit does not exist.
a) limx→−1−f(x) = 
b) limx→−1+f(x) = 
c) limx→−1f(x) = 
d) f(−1) = 
e) limx→1−f(x) = 
f) limx→1+f(x) = 
g) limx→1f(x) = 
h) f(1) = 

Learn how to evaluate limits for a piecewise function with discontinuities. This step-by-step guide covers left-hand and right-hand limits, function evaluation at specific points, and determining where limits do not exist. Suitable for Grade 12 or first-year college calculus students.

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