Calc Limits Worksheet Key and Solutions | PDF - Free Printable

Let's go through the worksheet on limits step by step, analyzing both problems based on the given graphs and functions.

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Problem 1: Piecewise Function with Graph

We are given a piecewise function:

$$
f(x) =
\begin{cases}
\frac{x+2}{x}, & x < 0 \\
|x-2|, & 0 \leq x < 3 \\
3, & x = 3 \\
-(x - 5), & x > 3
\end{cases}
$$

And we are to use the graph (which is sketched) to answer questions about values and limits.

Let’s analyze each part carefully.

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#### (a) $ f(0) = ? $

From the piecewise definition:
- For $ x = 0 $, it falls in the interval $ 0 \leq x < 3 $, so we use $ f(x) = |x - 2| $
- So $ f(0) = |0 - 2| = |-2| = 2 $

But wait — looking at the graph, there's a hole or discontinuity at $ x = 0 $. Let’s check what the graph shows.

Looking at the handwritten note:
> "Infinite Disc, x=0" → Infinite discontinuity at $ x = 0 $
> "Jump Disc, x=3" → Jump discontinuity at $ x = 3 $

Also, the graph shows a vertical asymptote at $ x = 0 $ from the left side because of $ \frac{x+2}{x} $, which blows up as $ x \to 0^- $.

But for $ x = 0 $, since $ x = 0 $ is not in the domain of the first piece ($ x < 0 $), and the second piece starts at $ x = 0 $, but let's look closely.

Wait: the function is defined as:
- $ \frac{x+2}{x} $ for $ x < 0 $
- $ |x - 2| $ for $ 0 \leq x < 3 $
- So at $ x = 0 $, it's defined by the second piece: $ f(0) = |0 - 2| = 2 $

BUT the graph shows a vertical asymptote at $ x = 0 $? That contradicts unless...

Wait! The handwritten note says "Infinite Disc, x=0", suggesting that $ f(0) $ might not be defined.

But according to the piecewise definition, $ f(0) = |0 - 2| = 2 $, so it should be defined.

However, if the graph shows a vertical asymptote at $ x = 0 $, then maybe there's a mistake in the piecewise definition?

Wait — look again at the function:

$$
f(x) =
\begin{cases}
\frac{x+2}{x}, & x < 0 \\
|x-2|, & 0 < x < 3 \\
3, & x = 3 \\
-(x - 5), & x > 3
\end{cases}
$$

Ah! There's a possible typo in the original write-up. It says:

> $ f(x) = \left\{ \frac{x+2}{x}, x<0 \right\} \left\{ |x-2|, 0<x<3 \right\} $

So $ x = 0 $ is NOT included in any piece?

But then $ f(0) $ is undefined.

Yes — that makes sense now.

The function is defined as:
- $ \frac{x+2}{x} $ for $ x < 0 $
- $ |x - 2| $ for $ 0 < x < 3 $
- $ f(3) = 3 $
- $ -(x - 5) $ for $ x > 3 $

So $ x = 0 $ is not in the domain → $ f(0) $ is undefined.

Hence:
> (a) $ f(0) = DNE $ ✔

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#### (b) $ f(2) = ? $

At $ x = 2 $, it's in $ 0 < x < 3 $, so use $ f(x) = |x - 2| $

So $ f(2) = |2 - 2| = 0 $

✔ Answer: $ f(2) = 0 $

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#### (c) $ f(3) = ? $

Given explicitly: $ f(3) = 3 $

✔ Answer: $ f(3) = 3 $

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#### (d) $ \lim_{x \to 0^-} f(x) = ? $

As $ x \to 0^- $, we’re approaching from the left, so use $ f(x) = \frac{x+2}{x} $

$$
\lim_{x \to 0^-} \frac{x+2}{x} = \frac{0 + 2}{0^-} = \frac{2}{\text{small negative}} \to -\infty
$$

So limit is $ -\infty $

✔ Answer: $ -\infty $

But handwritten says DNE (-∞) — correct, because $ -\infty $ is not a real number, so the limit does not exist (in finite terms), but we write $ -\infty $ when appropriate.

So: $ \boxed{-\infty} $ or $ \boxed{\text{DNE}} $? Since the instructions say "use $ \infty, -\infty $ or DNE", and this goes to $ -\infty $, we can write $ -\infty $

But sometimes teachers prefer DNE if infinite.

But here, the instruction says to use $ \infty $ or $ -\infty $ where appropriate, so we can use $ -\infty $.

✔ Answer: $ -\infty $

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#### (e) $ \lim_{x \to 0^+} f(x) = ? $

Approaching from the right: $ x \to 0^+ $, so use $ f(x) = |x - 2| $

As $ x \to 0^+ $, $ |x - 2| \to |0 - 2| = 2 $

So limit is $ 2 $

✔ Answer: $ 2 $

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#### (f) $ \lim_{x \to 3^-} f(x) = ? $

Approaching $ x = 3 $ from the left, so use $ f(x) = |x - 2| $ for $ x < 3 $

As $ x \to 3^- $, $ |x - 2| \to |3 - 2| = 1 $

So $ \lim_{x \to 3^-} f(x) = 1 $

But the handwritten says 2 — that’s incorrect.

Wait — let’s double-check.

Is $ |x - 2| $ defined for $ x \to 3^- $? Yes, and $ |3 - 2| = 1 $

But look at the graph: does it show a value of 2?

No — the graph shows a jump at $ x = 3 $.

Wait — perhaps the piecewise function is misread.

Wait — the function is:
- $ |x - 2| $ for $ 0 < x < 3 $
- $ f(3) = 3 $
- $ -(x - 5) $ for $ x > 3 $

So as $ x \to 3^- $, $ f(x) = |x - 2| \to |3 - 2| = 1 $

As $ x \to 3^+ $, $ f(x) = -(x - 5) = -x + 5 \to -3 + 5 = 2 $

And $ f(3) = 3 $

So:
- Left-hand limit: $ 1 $
- Right-hand limit: $ 2 $
- Value at 3: $ 3 $

So:
- $ \lim_{x \to 3^-} f(x) = 1 $
- $ \lim_{x \to 3^+} f(x) = 2 $
- $ f(3) = 3 $

But handwritten says $ \lim_{x \to 3^-} f(x) = 2 $ — that’s wrong

So correction:

(f) $ \lim_{x \to 3^-} f(x) = 1 $

But the student wrote 2 — likely mistake.

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#### (g) $ \lim_{x \to 3^+} f(x) = ? $

Use $ f(x) = -(x - 5) = -x + 5 $

As $ x \to 3^+ $, $ -x + 5 \to -3 + 5 = 2 $

So $ \lim_{x \to 3^+} f(x) = 2 $

✔ Answer: $ 2 $

But handwritten says DNE — that’s wrong. The right-hand limit exists and is 2.

So student made a mistake here.

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#### (h) $ \lim_{x \to \infty} f(x) = ? $

As $ x \to \infty $, $ f(x) = -(x - 5) = -x + 5 \to -\infty $

So $ \lim_{x \to \infty} f(x) = -\infty $

But handwritten says 1 — that’s completely wrong.

Wait — is it $ \lim_{x \to \infty} f(x) $ or $ \lim_{x \to -\infty} $?

Let’s check:

For $ x \to \infty $: $ f(x) = -(x - 5) = -x + 5 \to -\infty $

For $ x \to -\infty $: $ f(x) = \frac{x+2}{x} = 1 + \frac{2}{x} \to 1 $

So:
- $ \lim_{x \to -\infty} f(x) = 1 $
- $ \lim_{x \to \infty} f(x) = -\infty $

But the question says $ \lim_{x \to \infty} f(x) $, so it should be $ -\infty $

But handwritten says 1 — so they must have meant $ \lim_{x \to -\infty} f(x) $

Possibility: typo in the question or handwriting.

But the written question says: $ \lim_{x \to \infty} f(x) $

So correct answer is $ -\infty $

But student wrote 1 — that’s wrong.

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✔ Corrected Answers for Problem 1:

| Question | Answer |
|--------|--------|
| (a) $ f(0) $ | DNE |
| (b) $ f(2) $ | 0 |
| (c) $ f(3) $ | 3 |
| (d) $ \lim_{x \to 0^-} f(x) $ | $ -\infty $ |
| (e) $ \lim_{x \to 0^+} f(x) $ | 2 |
| (f) $ \lim_{x \to 3^-} f(x) $ | 1 |
| (g) $ \lim_{x \to 3^+} f(x) $ | 2 |
| (h) $ \lim_{x \to \infty} f(x) $ | $ -\infty $ |

Student's errors:
- (f): wrote 2 instead of 1
- (g): wrote DNE instead of 2
- (h): wrote 1 instead of $ -\infty $

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Problem 2: Rational Function

$$
f(x) = \frac{x(x - 3)}{(x + 1)(x - 2)}
$$

Graph has vertical asymptotes at $ x = -1 $ and $ x = 2 $, and a zero at $ x = 0 $, $ x = 3 $

Horizontal asymptote: degree numerator = degree denominator = 2 → ratio of leading coefficients: $ \frac{1}{1} = 1 $

So $ \lim_{x \to \pm\infty} f(x) = 1 $

Now evaluate:

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#### (a) $ f(0) = ? $

Plug in $ x = 0 $:

$$
f(0) = \frac{0(-3)}{(1)(-2)} = \frac{0}{-2} = 0
$$

✔ Answer: 0

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#### (b) $ f(2) = ? $

Denominator: $ (2 + 1)(2 - 2) = (3)(0) = 0 $, numerator: $ 2(-1) = -2 \neq 0 $

So undefined → DNE

✔ Answer: DNE

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#### (c) $ f(3) = ? $

$$
f(3) = \frac{3(0)}{(4)(1)} = 0
$$

✔ Answer: 0

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#### (d) $ \lim_{x \to -1} f(x) = ? $

Vertical asymptote at $ x = -1 $

Check left and right:

- As $ x \to -1^- $: $ x < -1 $
- Numerator: $ x(x - 3) $: negative × negative = positive
- Denominator: $ (x + 1) \to 0^- $, $ (x - 2) \to -3 $ → overall denominator → $ 0^- \times (-3) = 0^+ $
- So $ f(x) \to +\infty $

- As $ x \to -1^+ $: $ x > -1 $, close to -1
- $ x + 1 \to 0^+ $, $ x - 2 \to -3 $
- Denominator → $ 0^+ \times (-3) = 0^- $
- Numerator: $ x(x - 3) \approx (-1)(-4) = 4 > 0 $
- So $ f(x) \to -\infty $

Since LHL ≠ RHL, limit does not exist

✔ Answer: DNE

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#### (e) $ \lim_{x \to 0} f(x) = ? $

$ f(x) $ is continuous at $ x = 0 $, since no division by zero

We already computed $ f(0) = 0 $

So $ \lim_{x \to 0} f(x) = 0 $

✔ Answer: 0

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#### (f) $ \lim_{x \to 2^+} f(x) = ? $

Approach $ x = 2 $ from the right

- $ x \to 2^+ $: $ x > 2 $
- $ x + 1 \to 3 $, $ x - 2 \to 0^+ $
- Numerator: $ x(x - 3) \to 2(-1) = -2 $
- Denominator: $ (positive)(small positive) \to 0^+ $
- So $ f(x) \to \frac{-2}{0^+} = -\infty $

But student wrote DNE (-∞) — acceptable, but usually we write $ -\infty $

So answer is $ -\infty $

But student wrote DNE (-∞) — okay, but could just write $ -\infty $

✔ Answer: $ -\infty $

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#### (g) $ \lim_{x \to \infty} f(x) = ? $

As $ x \to \infty $, leading term: $ \frac{x^2}{x^2} = 1 $

So $ \lim_{x \to \infty} f(x) = 1 $

✔ Answer: 1

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✔ Corrected Answers for Problem 2:

| Question | Answer |
|--------|--------|
| (a) $ f(0) $ | 0 |
| (b) $ f(2) $ | DNE |
| (c) $ f(3) $ | 0 |
| (d) $ \lim_{x \to -1} f(x) $ | DNE |
| (e) $ \lim_{x \to 0} f(x) $ | 0 |
| (f) $ \lim_{x \to 2^+} f(x) $ | $ -\infty $ |
| (g) $ \lim_{x \to \infty} f(x) $ | 1 |

All answers match the student’s except possibly (f), where student wrote DNE (-∞) — which is acceptable if they mean the limit doesn't exist (infinite), but technically $ -\infty $ is fine.

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🔍 Summary of Corrections:

#### Problem 1 Errors:
- (f) $ \lim_{x \to 3^-} f(x) = 1 $, not 2
- (g) $ \lim_{x \to 3^+} f(x) = 2 $, not DNE
- (h) $ \lim_{x \to \infty} f(x) = -\infty $, not 1

#### Problem 2:
- All correct, except (f): $ -\infty $ is better than DNE(-∞)

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✔ Final Correct Answers:

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Problem 1:

(a) $ f(0) = \boxed{\text{DNE}} $
(b) $ f(2) = \boxed{0} $
(c) $ f(3) = \boxed{3} $
(d) $ \lim_{x \to 0^-} f(x) = \boxed{-\infty} $
(e) $ \lim_{x \to 0^+} f(x) = \boxed{2} $
(f) $ \lim_{x \to 3^-} f(x) = \boxed{1} $
(g) $ \lim_{x \to 3^+} f(x) = \boxed{2} $
(h) $ \lim_{x \to \infty} f(x) = \boxed{-\infty} $

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Problem 2:

(a) $ f(0) = \boxed{0} $
(b) $ f(2) = \boxed{\text{DNE}} $
(c) $ f(3) = \boxed{0} $
(d) $ \lim_{x \to -1} f(x) = \boxed{\text{DNE}} $
(e) $ \lim_{x \to 0} f(x) = \boxed{0} $
(f) $ \lim_{x \to 2^+} f(x) = \boxed{-\infty} $
(g) $ \lim_{x \to \infty} f(x) = \boxed{1} $

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Let me know if you'd like a visual explanation or graph analysis!