Tricky Limit Problems worksheet featuring graphs of functions f(x), g(x), h(x), and j(x) for evaluating limits.

Let's solve each of these limit problems step by step using the provided graphs of $ f(x), g(x), h(x), $ and $ j(x) $. We will analyze each function based on its graph and evaluate the limits accordingly.

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Graphs Summary (based on visual interpretation):

- $ y = f(x) $:
- A piecewise constant function.
- For $ x < 0 $, $ f(x) = -2 $.
- At $ x = 0 $, there is a hole (open circle), so $ f(0) $ is undefined.
- For $ x > 0 $, $ f(x) = 3 $.
- So:
$$
\lim_{x \to 0^-} f(x) = -2,\quad \lim_{x \to 0^+} f(x) = 3
$$

- $ y = g(x) $:
- A V-shaped graph (like absolute value), but shifted.
- It appears to be linear with a peak at $ x = 1 $, where $ g(1) = 3 $.
- Left side: from $ (-1, 1) $ to $ (1, 3) $ → slope = 1
- Right side: from $ (1, 3) $ to $ (3, 1) $ → slope = -1
- So $ g(x) = |x - 1| + 2 $? Wait, let’s verify:
- At $ x = 1 $: $ g(1) = 3 $ ✔
- At $ x = 0 $: $ g(0) = 2 $? But graph shows $ g(0) = 2 $? Actually, point at $ (0, 2) $, $ (1, 3) $, $ (2, 2) $, $ (3, 1) $ → seems like $ g(x) = -|x - 1| + 3 $?
- $ x = 0 $: $ -| -1 | + 3 = -1 + 3 = 2 $ ✔
- $ x = 1 $: $ -0 + 3 = 3 $ ✔
- $ x = 2 $: $ -1 + 3 = 2 $ ✔
- $ x = 3 $: $ -2 + 3 = 1 $ ✔
- So $ g(x) = -|x - 1| + 3 $

- $ y = h(x) $:
- A curve that passes through $ (-2, -2) $, goes up to $ (-1, 2) $, then has an open circle at $ x = 0 $, and continues from $ (0, -1) $ to $ (1, 0) $, then $ (2, 2) $.
- There is a jump discontinuity at $ x = 0 $.
- From left: $ h(x) \to 2 $ as $ x \to 0^- $
- From right: $ h(x) \to -1 $ as $ x \to 0^+ $
- So $ \lim_{x \to 0^-} h(x) = 2 $, $ \lim_{x \to 0^+} h(x) = -1 $

- $ y = j(x) $:
- Has vertical asymptotes at $ x = -2 $ and $ x = 2 $.
- As $ x \to -2^- $, $ j(x) \to -\infty $
- As $ x \to -2^+ $, $ j(x) \to +\infty $
- As $ x \to 2^- $, $ j(x) \to -\infty $
- As $ x \to 2^+ $, $ j(x) \to +\infty $
- Also, $ j(0) = 0 $, $ j(-4) = -2 $, $ j(4) = 2 $, etc.
- Horizontal asymptote: $ y = 2 $ as $ x \to \pm\infty $

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Now, let’s solve each problem:

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1. $ \displaystyle \lim_{x \to -2} j(x) = ? $

From the graph:
- As $ x \to -2^- $, $ j(x) \to -\infty $
- As $ x \to -2^+ $, $ j(x) \to +\infty $
- Since left and right limits are not equal and both go to infinity in opposite directions, the limit does not exist.

But wait — the question says "solve" and the answer might be written as $ \infty $ or DNE.

However, since one side goes to $ -\infty $ and the other to $ +\infty $, the two-sided limit does not exist.

> ✔ Answer: DNE

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2. $ \displaystyle \lim_{x \to 1} j(x) = ? $

Look at $ j(x) $ near $ x = 1 $:
- The graph is continuous around $ x = 1 $.
- $ j(1) = 0 $, and it's smooth there.
- So $ \lim_{x \to 1} j(x) = j(1) = 0 $

> ✔ Answer: 0

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3. $ \displaystyle \lim_{x \to 1} \frac{f(x) - 2}{f(x)} = ? $

We need to evaluate this expression as $ x \to 1 $. First, find $ f(1) $.

From $ f(x) $ graph:
- For $ x > 0 $, $ f(x) = 3 $
- So $ f(1) = 3 $

Thus:
$$
\frac{f(x) - 2}{f(x)} \to \frac{3 - 2}{3} = \frac{1}{3}
$$

Since $ f(x) $ is constant for $ x > 0 $, and $ x = 1 > 0 $, the limit exists and equals $ \frac{1}{3} $

> ✔ Answer: $ \frac{1}{3} $

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4. $ \displaystyle \lim_{x \to 0} h(f(x)) = ? $

This is a composition: $ h(f(x)) $

First, determine what $ f(x) $ approaches as $ x \to 0 $.

From earlier:
- $ \lim_{x \to 0^-} f(x) = -2 $
- $ \lim_{x \to 0^+} f(x) = 3 $

So we must check left and right limits separately.

#### Left-hand limit ($ x \to 0^- $):
- $ f(x) \to -2 $
- So $ h(f(x)) \to h(-2) $
- From $ h(x) $ graph: $ h(-2) = -2 $
- So $ \lim_{x \to 0^-} h(f(x)) = h(-2) = -2 $

#### Right-hand limit ($ x \to 0^+ $):
- $ f(x) \to 3 $
- So $ h(f(x)) \to h(3) $
- From $ h(x) $ graph: $ h(3) = 2 $
- So $ \lim_{x \to 0^+} h(f(x)) = 2 $

Left and right limits differ: $ -2 $ vs $ 2 $

> ✔ Answer: DNE

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5. $ \displaystyle \lim_{x \to 1} f(g(x)) = ? $

Again, composition: $ f(g(x)) $

First, find $ g(1) $:
- $ g(1) = 3 $ (from graph)

Now, as $ x \to 1 $, $ g(x) \to 3 $, since $ g $ is continuous at $ x = 1 $

Then $ f(g(x)) \to f(3) $

Now, $ f(3) $: since $ 3 > 0 $, $ f(3) = 3 $

So:
$$
\lim_{x \to 1} f(g(x)) = f(g(1)) = f(3) = 3
$$

> ✔ Answer: 3

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6. $ \displaystyle \lim_{x \to 0} f(|x| + 2) = ? $

Let’s define $ u = |x| + 2 $

As $ x \to 0 $, $ |x| \to 0 $, so $ u \to 2 $

So $ f(|x| + 2) \to f(2) $

Now, $ f(2) $: since $ 2 > 0 $, $ f(2) = 3 $

So:
$$
\lim_{x \to 0} f(|x| + 2) = f(2) = 3
$$

> ✔ Answer: 3

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7. $ \displaystyle \lim_{x \to 1} \left( g(x) \cdot f(x + 2) \right) = ? $

Evaluate $ g(x) \cdot f(x + 2) $ as $ x \to 1 $

First, $ g(1) = 3 $, and $ g $ is continuous at $ x = 1 $

Now, $ f(x + 2) $: when $ x \to 1 $, $ x + 2 \to 3 $

So $ f(x + 2) \to f(3) = 3 $ (since $ 3 > 0 $)

So:
$$
\lim_{x \to 1} g(x) \cdot f(x + 2) = g(1) \cdot f(3) = 3 \cdot 3 = 9
$$

> ✔ Answer: 9

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8. $ \displaystyle \lim_{x \to 2} \left( (f(x) - 1)^2 - 6 \right) = ? $

First, $ f(x) $ at $ x = 2 $: $ f(2) = 3 $

So $ f(x) \to 3 $ as $ x \to 2 $ (since $ x > 0 $, $ f(x) = 3 $)

Then:
$$
(f(x) - 1)^2 - 6 \to (3 - 1)^2 - 6 = 4 - 6 = -2
$$

> ✔ Answer: -2

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9. $ \displaystyle \lim_{x \to 2} \left( h(x) + f(x) \right) = ? $

Check $ h(x) $ and $ f(x) $ as $ x \to 2 $

- $ h(2) = 2 $ (from graph)
- $ f(2) = 3 $
- Both functions are continuous at $ x = 2 $, so:
$$
\lim_{x \to 2} (h(x) + f(x)) = h(2) + f(2) = 2 + 3 = 5
$$

> ✔ Answer: 5

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10. $ \displaystyle \lim_{x \to -2} j(f(x)) = ? $

We need $ j(f(x)) $ as $ x \to -2 $

First, find $ f(x) $ as $ x \to -2 $

Since $ -2 < 0 $, $ f(x) = -2 $ for all $ x < 0 $

So $ f(x) \to -2 $ as $ x \to -2 $

Then $ j(f(x)) \to j(-2) $

But look at $ j(x) $: at $ x = -2 $, there is a vertical asymptote

As $ x \to -2 $, $ j(x) \to \pm\infty $

But here, we're evaluating $ j(f(x)) $, and $ f(x) \to -2 $, so we are evaluating $ j $ at values approaching $ -2 $

So $ j(f(x)) \to j(-2) $, but $ j $ has a vertical asymptote at $ x = -2 $

We need to know the behavior of $ j(x) $ as $ x \to -2 $

From graph:
- As $ x \to -2^- $, $ j(x) \to -\infty $
- As $ x \to -2^+ $, $ j(x) \to +\infty $

But $ f(x) = -2 $ for all $ x < 0 $, so $ f(x) \to -2 $ from the left? Wait:

Wait! $ f(x) = -2 $ for all $ x < 0 $, so $ f(x) $ is constantly $ -2 $ as $ x \to -2 $

So $ f(x) = -2 $, so $ j(f(x)) = j(-2) $

But $ j(-2) $ is undefined, because there's a vertical asymptote at $ x = -2 $

Moreover, $ j(-2) $ is not defined, and as $ x \to -2 $, $ j(x) $ blows up.

But since $ f(x) = -2 $ exactly for $ x < 0 $, and $ x \to -2 $, $ f(x) = -2 $, so we’re plugging $ -2 $ into $ j $, which is undefined.

So $ j(f(x)) = j(-2) $, which is undefined.

But is the limit still possible?

Wait: as $ x \to -2 $, $ f(x) = -2 $, so $ j(f(x)) = j(-2) $, which is not defined, and the function $ j $ has no value at $ x = -2 $

Therefore, $ j(f(x)) $ is undefined at $ x \to -2 $, and since $ f(x) = -2 $, and $ j(-2) $ is undefined, the expression $ j(f(x)) $ is undefined near $ x = -2 $?

Wait: actually, $ f(x) = -2 $ for $ x < 0 $, so $ f(x) $ is constantly $ -2 $, so $ j(f(x)) = j(-2) $, which is undefined.

But in the limit, we care about values near $ x = -2 $, not necessarily at $ x = -2 $

But since $ f(x) = -2 $ for all $ x < 0 $, then $ j(f(x)) = j(-2) $, which is undefined for all $ x < 0 $ near $ -2 $

So $ j(f(x)) $ is not defined in any neighborhood of $ x = -2 $, so the limit does not exist (because the function isn't defined near $ x = -2 $)

Alternatively, if we consider the domain, $ j(-2) $ is undefined, so $ j(f(x)) $ is undefined for all $ x < 0 $, hence the expression is undefined near $ x = -2 $, so the limit does not exist.

> ✔ Answer: DNE

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11. $ \displaystyle \lim_{x \to 1.5} \frac{g(x) - 1}{2x - 3} = ? $

First, plug in $ x = 1.5 $

- $ g(1.5) $: from graph, $ g(x) = -|x - 1| + 3 $
- $ g(1.5) = -|0.5| + 3 = -0.5 + 3 = 2.5 $
- So numerator: $ g(1.5) - 1 = 2.5 - 1 = 1.5 $
- Denominator: $ 2(1.5) - 3 = 3 - 3 = 0 $

So we have $ \frac{1.5}{0} $ — division by zero!

But we need to see the limit, not just the value.

Let’s compute:
$$
\lim_{x \to 1.5} \frac{g(x) - 1}{2x - 3}
$$

Note that $ 2x - 3 = 0 $ at $ x = 1.5 $, so denominator approaches 0.

Let’s use the definition of $ g(x) $:

$ g(x) = -|x - 1| + 3 $

At $ x = 1.5 $, $ x > 1 $, so $ |x - 1| = x - 1 $, so:
$$
g(x) = -(x - 1) + 3 = -x + 1 + 3 = -x + 4
$$

So for $ x > 1 $, $ g(x) = -x + 4 $

Then:
$$
g(x) - 1 = -x + 4 - 1 = -x + 3
$$

Denominator: $ 2x - 3 $

So:
$$
\frac{g(x) - 1}{2x - 3} = \frac{-x + 3}{2x - 3}
$$

Now take limit as $ x \to 1.5 $:
$$
\frac{-(1.5) + 3}{2(1.5) - 3} = \frac{1.5}{0} \Rightarrow \text{undefined}
$$

But let's check left and right:

For $ x \to 1.5^- $: $ x < 1.5 $, but still $ x > 1 $, so same formula applies: $ g(x) = -x + 4 $

So expression is $ \frac{-x + 3}{2x - 3} $

As $ x \to 1.5^- $, numerator $ \to 1.5 $, denominator $ \to 0^- $ (since $ 2x - 3 < 0 $ for $ x < 1.5 $)

So $ \frac{1.5}{0^-} \to -\infty $

As $ x \to 1.5^+ $, denominator $ \to 0^+ $, so $ \frac{1.5}{0^+} \to +\infty $

So left limit: $ -\infty $, right limit: $ +\infty $

Therefore, two-sided limit does not exist

> ✔ Answer: DNE

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12. $ \displaystyle \lim_{x \to 0} g(f(x) + 1) = ? $

We want $ g(f(x) + 1) $ as $ x \to 0 $

First, $ f(x) $:
- As $ x \to 0^- $, $ f(x) \to -2 $
- As $ x \to 0^+ $, $ f(x) \to 3 $

So $ f(x) + 1 $:
- $ x \to 0^- $: $ f(x) + 1 \to -2 + 1 = -1 $
- $ x \to 0^+ $: $ f(x) + 1 \to 3 + 1 = 4 $

Now apply $ g $:
- $ g(-1) $: from graph, $ g(-1) = 2 $ (since $ g(x) = -|x - 1| + 3 $, $ g(-1) = -|-2| + 3 = -2 + 3 = 1 $? Wait — contradiction.)

Wait, let’s recheck $ g(x) $ values.

Earlier we said $ g(x) = -|x - 1| + 3 $

- $ x = -1 $: $ |-1 - 1| = 2 $, so $ g(-1) = -2 + 3 = 1 $
- But from graph: point at $ (-1, 1) $? Yes — yes, graph shows $ (-1, 1) $

So $ g(-1) = 1 $

Similarly:
- $ g(4) = -|4 - 1| + 3 = -3 + 3 = 0 $

But $ f(x) + 1 \to -1 $ as $ x \to 0^- $, so $ g(f(x) + 1) \to g(-1) = 1 $

As $ x \to 0^+ $, $ f(x) + 1 \to 4 $, so $ g(f(x) + 1) \to g(4) = 0 $

So:
- Left limit: $ 1 $
- Right limit: $ 0 $

Not equal ⇒ limit does not exist

> ✔ Answer: DNE

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✔ Final Answers:

1. $ \boxed{\text{DNE}} $
2. $ \boxed{0} $
3. $ \boxed{\frac{1}{3}} $
4. $ \boxed{\text{DNE}} $
5. $ \boxed{3} $
6. $ \boxed{3} $
7. $ \boxed{9} $
8. $ \boxed{-2} $
9. $ \boxed{5} $
10. $ \boxed{\text{DNE}} $
11. $ \boxed{\text{DNE}} $
12. $ \boxed{\text{DNE}} $

Let me know if you'd like a diagram explanation or clarification on any step!