Math Exercises & Math Problems: Limit of a Function - Free Printable

It looks like you've uploaded a list of limit problems (labeled a) through z)), but no image was actually attached in your message. However, I can still help you solve these limits if you provide the expressions, or if they're already written here.

Looking at your message, it appears that the limits are already typed out clearly — so let's go ahead and solve each one step by step, explaining the reasoning.

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We’ll evaluate each limit using algebraic simplification, factoring, cancellation, and standard limit rules (like direct substitution, L’Hôpital’s Rule when needed). Let's proceed one by one.

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a) $\lim_{x \to 2} (x^2 - 4)$

Direct substitution:
$$
(2)^2 - 4 = 4 - 4 = 0
$$
✔ Answer: $0$

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b) $\lim_{x \to 0} \frac{x^3 - 4x}{2x^2 + 3x}$

Factor numerator and denominator:
Numerator: $x(x^2 - 4) = x(x - 2)(x + 2)$
Denominator: $x(2x + 3)$

Cancel $x$ (since $x \to 0$, $x \neq 0$):
$$
\frac{x(x - 2)(x + 2)}{x(2x + 3)} = \frac{(x - 2)(x + 2)}{2x + 3}
$$
Now substitute $x = 0$:
$$
\frac{(-2)(2)}{3} = \frac{-4}{3}
$$
✔ Answer: $-\frac{4}{3}$

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c) $\lim_{x \to -1} \frac{x^3}{(x+1)^2}$

Try substitution: $x = -1$
Numerator: $(-1)^3 = -1$
Denominator: $(0)^2 = 0$

So we have $\frac{-1}{0}$ → undefined, but check sign.

As $x \to -1^-$: $x < -1$, so $x+1 < 0$, $(x+1)^2 > 0$, numerator $< 0$: negative over positive → $-\infty$

As $x \to -1^+$: $x > -1$, $x+1 > 0$, $(x+1)^2 > 0$, numerator $< 0$: still negative over positive → $-\infty$

So both sides go to $-\infty$

✔ Answer: $-\infty$

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d) $\lim_{x \to -1} \frac{(x+1)^2(x-1)}{x^3 + 1}$

Note: $x^3 + 1 = (x + 1)(x^2 - x + 1)$

Numerator: $(x+1)^2(x - 1)$
Denominator: $(x + 1)(x^2 - x + 1)$

Cancel $x+1$ (valid since $x \ne -1$):
$$
\frac{(x+1)(x - 1)}{x^2 - x + 1}
$$

Now plug in $x = -1$:
Numerator: $(-1 + 1)(-1 - 1) = (0)(-2) = 0$
Denominator: $1 + 1 + 1 = 3$

✔ Answer: $0$

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e) $\lim_{x \to 0} \frac{x^3 - 2x^2 + x}{2x^3 + x^2 - 2x}$

Factor numerator and denominator:

Numerator: $x(x^2 - 2x + 1) = x(x - 1)^2$
Denominator: $x(2x^2 + x - 2)$

Factor $2x^2 + x - 2$:
Discriminant: $1 + 16 = 17$, roots: $x = \frac{-1 \pm \sqrt{17}}{4}$ → not nice.

But cancel $x$ (since $x \to 0$, $x \ne 0$):

$$
\frac{(x - 1)^2}{2x^2 + x - 2}
$$

Now plug in $x = 0$:
Numerator: $1$
Denominator: $-2$

✔ Answer: $-\frac{1}{2}$

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

Substitute $x = 1$:
Numerator: $1 + 2 + 3 = 6$
Denominator: $0^2 = 0$

So $\frac{6}{0}$ → undefined.

Check sign: $(x - 1)^2 > 0$ for all $x \ne 1$, so denominator always positive.

Numerator → 6 > 0 ⇒ expression → $+\infty$

✔ Answer: $+\infty$

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g) $\lim_{x \to 0} \frac{x^4 - 4x^3 + x^2}{x^3 + x^2 + x}$

Factor numerator and denominator:

Numerator: $x^2(x^2 - 4x + 1)$
Denominator: $x(x^2 + x + 1)$

Cancel $x$ (since $x \to 0$, $x \ne 0$):

$$
\frac{x(x^2 - 4x + 1)}{x^2 + x + 1}
$$

Now plug in $x = 0$:
Numerator: $0$
Denominator: $1$

✔ Answer: $0$

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h) $\lim_{x \to -1} \frac{x^3 + x^2 + x + 1}{x^4 + x^2 - 2}$

Try factoring:

Numerator: group: $(x^3 + x^2) + (x + 1) = x^2(x + 1) + 1(x + 1) = (x^2 + 1)(x + 1)$

Denominator: $x^4 + x^2 - 2$. Let $u = x^2$: $u^2 + u - 2 = (u + 2)(u - 1)$ → $(x^2 + 2)(x^2 - 1) = (x^2 + 2)(x - 1)(x + 1)$

So full expression:
$$
\frac{(x^2 + 1)(x + 1)}{(x^2 + 2)(x - 1)(x + 1)}
$$

Cancel $x + 1$ (since $x \to -1$, $x \ne -1$):

$$
\frac{x^2 + 1}{(x^2 + 2)(x - 1)}
$$

Now plug in $x = -1$:
Numerator: $1 + 1 = 2$
Denominator: $(1 + 2)(-1 - 1) = 3 \cdot (-2) = -6$

✔ Answer: $-\frac{1}{3}$

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i) $\lim_{x \to 2} \frac{(x + 1)^2}{2 - x}$

Plug in $x = 2$:
Numerator: $(3)^2 = 9$
Denominator: $2 - 2 = 0$

So $\frac{9}{0}$ → check sign.

As $x \to 2^-$: $x < 2$, so $2 - x > 0$ → $+\infty$
As $x \to 2^+$: $x > 2$, $2 - x < 0$ → $-\infty$

One-sided limits differ → limit does not exist

✔ Answer: DNE (does not exist)

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j) $\lim_{x \to 2} \frac{x - 2}{x^2 - 3x + 2}$

Factor denominator: $x^2 - 3x + 2 = (x - 1)(x - 2)$

So:
$$
\frac{x - 2}{(x - 1)(x - 2)} = \frac{1}{x - 1}, \quad x \ne 2
$$

Now take limit as $x \to 2$: $\frac{1}{2 - 1} = 1$

✔ Answer: $1$

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k) $\lim_{x \to 0} \frac{3x + 2x^{-1}}{x + 4x^{-1}}$

Rewrite with $x^{-1} = \frac{1}{x}$:

$$
\frac{3x + \frac{2}{x}}{x + \frac{4}{x}} = \frac{\frac{3x^2 + 2}{x}}{\frac{x^2 + 4}{x}} = \frac{3x^2 + 2}{x^2 + 4}
$$

Now take $x \to 0$:
Numerator → $2$, Denominator → $4$

✔ Answer: $\frac{1}{2}$

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l) $\lim_{x \to 2} \frac{x^2 - 3x + 2}{x^2 - 2x}$

Factor:
Numerator: $(x - 1)(x - 2)$
Denominator: $x(x - 2)$

Cancel $x - 2$ (since $x \to 2$, $x \ne 2$):

$$
\frac{x - 1}{x}
$$

Now plug in $x = 2$: $\frac{1}{2}$

✔ Answer: $\frac{1}{2}$

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m) $\lim_{x \to 1} \left(\frac{1}{1 - x} - \frac{3}{1 - x^3}\right)$

Note: $1 - x^3 = (1 - x)(1 + x + x^2)$

So rewrite:
$$
\frac{1}{1 - x} - \frac{3}{(1 - x)(1 + x + x^2)} = \frac{1}{1 - x} \left(1 - \frac{3}{1 + x + x^2}\right)
$$

Better: common denominator is $(1 - x)(1 + x + x^2)$

Write:
$$
\frac{1}{1 - x} = \frac{1 + x + x^2}{(1 - x)(1 + x + x^2)}
$$

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

Factor numerator: $x^2 + x - 2 = (x + 2)(x - 1)$

So:
$$
\frac{(x + 2)(x - 1)}{(1 - x)(1 + x + x^2)} = \frac{(x + 2)(x - 1)}{-(x - 1)(1 + x + x^2)} = \frac{x + 2}{-(1 + x + x^2)}, \quad x \ne 1
$$

Now plug in $x = 1$: $\frac{3}{-(1 + 1 + 1)} = \frac{3}{-3} = -1$

✔ Answer: $-1$

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n) $\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x}$

Factor:
Numerator: $(x - 2)(x + 1)$
Denominator: $x(x - 2)$

Cancel $x - 2$ (since $x \to 2$, $x \ne 2$):

$$
\frac{x + 1}{x} \to \frac{3}{2}
$$

✔ Answer: $\frac{3}{2}$

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o) $\lim_{x \to 2} \frac{x^2 + 5}{x^2 - 3}$

Plug in $x = 2$: $\frac{4 + 5}{4 - 3} = \frac{9}{1} = 9$

✔ Answer: $9$

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p) $\lim_{x \to 1} \frac{3x^4 - 4x^3 + 1}{(x - 1)^2}$

Try factoring numerator.

Let $f(x) = 3x^4 - 4x^3 + 1$

Try $x = 1$: $3 - 4 + 1 = 0$ → $x = 1$ is root.

Divide by $(x - 1)$:

Use polynomial division or synthetic division.

Synthetic division for root 1:

Coefficients: 3 | -4 | 0 | 0 | 1? Wait: degree 4 → $3x^4 - 4x^3 + 0x^2 + 0x + 1$

Wait, no: original is $3x^4 - 4x^3 + 1$, so coefficients: 3, -4, 0, 0, 1?

No! It's $3x^4 - 4x^3 + 1$ → missing $x^2$, $x$ terms → coeffs: 3, -4, 0, 0, 1

But $f(1) = 3 - 4 + 1 = 0$ → yes, root.

Synthetic division:

```
1 | 3 -4 0 0 1
| 3 -1 -1 -1
-----------------------
3 -1 -1 -1 0
```

So quotient: $3x^3 - x^2 - x - 1$

Now check if $x = 1$ is root again: $3 - 1 - 1 - 1 = 0$ → yes!

Divide again:

```
1 | 3 -1 -1 -1
| 3 2 1
-------------------
3 2 1 0
```

Quotient: $3x^2 + 2x + 1$

So overall:
$$
3x^4 - 4x^3 + 1 = (x - 1)^2(3x^2 + 2x + 1)
$$

So:
$$
\frac{(x - 1)^2(3x^2 + 2x + 1)}{(x - 1)^2} = 3x^2 + 2x + 1, \quad x \ne 1
$$

Now plug in $x = 1$: $3 + 2 + 1 = 6$

✔ Answer: $6$

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q) $\lim_{x \to -2} \frac{3x + 6}{x^3 + 8}$

Note: $x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)$

Numerator: $3(x + 2)$

So:
$$
\frac{3(x + 2)}{(x + 2)(x^2 - 2x + 4)} = \frac{3}{x^2 - 2x + 4}, \quad x \ne -2
$$

Now plug in $x = -2$: $(-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12$

✔ Answer: $\frac{3}{12} = \frac{1}{4}$

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r) $\lim_{x \to 2} \frac{x + 1}{x - 1}$

Plug in $x = 2$: $\frac{3}{1} = 3$

✔ Answer: $3$

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s) $\lim_{x \to -2} \frac{x^3 + 3x^2 + 2x}{x^2 - x - 6}$

Factor numerator: $x(x^2 + 3x + 2) = x(x + 1)(x + 2)$

Denominator: $x^2 - x - 6 = (x - 3)(x + 2)$

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

Now plug in $x = -2$: $\frac{(-2)(-1)}{-2 - 3} = \frac{2}{-5} = -\frac{2}{5}$

✔ Answer: $-\frac{2}{5}$

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t) $\lim_{x \to 1} \frac{x^2 - 2x + 1}{x^3 - x}$

Numerator: $(x - 1)^2$
Denominator: $x(x^2 - 1) = x(x - 1)(x + 1)$

So:
$$
\frac{(x - 1)^2}{x(x - 1)(x + 1)} = \frac{x - 1}{x(x + 1)}, \quad x \ne 1
$$

Now plug in $x = 1$: $\frac{0}{1 \cdot 2} = 0$

✔ Answer: $0$

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u) $\lim_{x \to 2} \frac{x^2 + 7x - 44}{x^2 - 6x + 8}$

Factor:

Numerator: $x^2 + 7x - 44$ → discriminant: $49 + 176 = 225 = 15^2$

Roots: $x = \frac{-7 \pm 15}{2}$ → $x = 4$, $x = -11$

So: $(x - 4)(x + 11)$

Denominator: $x^2 - 6x + 8 = (x - 2)(x - 4)$

So:
$$
\frac{(x - 4)(x + 11)}{(x - 2)(x - 4)} = \frac{x + 11}{x - 2}, \quad x \ne 4
$$

Now plug in $x = 2$: $\frac{13}{0}$ → undefined

Check sign: as $x \to 2^-$, $x - 2 < 0$ → $-\infty$
As $x \to 2^+$, $x - 2 > 0$ → $+\infty$

So limit does not exist

✔ Answer: DNE

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v) $\lim_{x \to 2} \frac{x^2 - 4}{x^2 - 3x + 2}$

Numerator: $(x - 2)(x + 2)$
Denominator: $(x - 1)(x - 2)$

Cancel $x - 2$ (since $x \to 2$, $x \ne 2$):

$$
\frac{x + 2}{x - 1} \to \frac{4}{1} = 4
$$

✔ Answer: $4$

---

w) $\lim_{x \to 1} \frac{x^3 - 5x + 4}{x^3 - 1}$

Numerator: $x^3 - 5x + 4$

Try $x = 1$: $1 - 5 + 4 = 0$ → root

Divide by $x - 1$:

Synthetic division:

```
1 | 1 0 -5 4
| 1 1 -4
-----------------
1 1 -4 0
```

So: $(x - 1)(x^2 + x - 4)$

Denominator: $x^3 - 1 = (x - 1)(x^2 + x + 1)$

So:
$$
\frac{(x - 1)(x^2 + x - 4)}{(x - 1)(x^2 + x + 1)} = \frac{x^2 + x - 4}{x^2 + x + 1}, \quad x \ne 1
$$

Now plug in $x = 1$: $\frac{1 + 1 - 4}{1 + 1 + 1} = \frac{-2}{3}$

✔ Answer: $-\frac{2}{3}$

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x) $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Numerator: $(x - 2)(x + 2)$

So:
$$
\frac{(x - 2)(x + 2)}{x - 2} = x + 2, \quad x \ne 2
$$

→ $x \to 2$: $2 + 2 = 4$

✔ Answer: $4$

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y) $\lim_{x \to 2} \frac{x^2 - 4}{x^2 - 3x + 2}$

This is same as v): already solved → $4$

✔ Answer: $4$

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z) $\lim_{x \to 1} \left(\frac{1}{x^2 - 1} - \frac{2}{x^4 - 1}\right)$

Note: $x^2 - 1 = (x - 1)(x + 1)$
$x^4 - 1 = (x^2)^2 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)$

So:
$$
\frac{1}{(x - 1)(x + 1)} - \frac{2}{(x - 1)(x + 1)(x^2 + 1)}
= \frac{1}{(x - 1)(x + 1)} \left(1 - \frac{2}{x^2 + 1}\right)
$$

Common denominator:
$$
\frac{(x^2 + 1) - 2}{(x - 1)(x + 1)(x^2 + 1)} = \frac{x^2 - 1}{(x - 1)(x + 1)(x^2 + 1)}
$$

But $x^2 - 1 = (x - 1)(x + 1)$, so:

$$
\frac{(x - 1)(x + 1)}{(x - 1)(x + 1)(x^2 + 1)} = \frac{1}{x^2 + 1}, \quad x \ne \pm 1
$$

Now plug in $x = 1$: $\frac{1}{1 + 1} = \frac{1}{2}$

✔ Answer: $\frac{1}{2}$

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Z) $\lim_{x \to 3} \frac{x - 3}{x^2 - 5x + 6}$

Factor denominator: $x^2 - 5x + 6 = (x - 2)(x - 3)$

So:
$$
\frac{x - 3}{(x - 2)(x - 3)} = \frac{1}{x - 2}, \quad x \ne 3
$$

Now plug in $x = 3$: $\frac{1}{1} = 1$

✔ Answer: $1$

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| a) | $0$ |
| b) | $-\frac{4}{3}$ |
| c) | $-\infty$ |
| d) | $0$ |
| e) | $-\frac{1}{2}$ |
| f) | $+\infty$ |
| g) | $0$ |
| h) | $-\frac{1}{3}$ |
| i) | DNE |
| j) | $1$ |
| k) | $\frac{1}{2}$ |
| l) | $\frac{1}{2}$ |
| m) | $-1$ |
| n) | $\frac{3}{2}$ |
| o) | $9$ |
| p) | $6$ |
| q) | $\frac{1}{4}$ |
| r) | $3$ |
| s) | $-\frac{2}{5}$ |
| t) | $0$ |
| u) | DNE |
| v) | $4$ |
| w) | $-\frac{2}{3}$ |
| x) | $4$ |
| y) | $4$ |
| z) | $\frac{1}{2}$ |
| Z) | $1$ |

Let me know if you'd like any specific steps explained further!