Math worksheet with limit problems requiring evaluation of expressions as variables approach specific values.

Let's solve each of these limit problems step by step. I'll go through them in order, explaining the reasoning and techniques used.

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1. Find the following limits:

#### (a) $\lim_{x \to 2} (2x^2 + 3x - 14)$

This is a polynomial function, so we can directly substitute $x = 2$:

$$
2(2)^2 + 3(2) - 14 = 2(4) + 6 - 14 = 8 + 6 - 14 = 0
$$

✔ Answer: $0$

---

#### (b) $\lim_{x \to 5} (x^2 + 2x - 9)$

Again, direct substitution:

$$
(5)^2 + 2(5) - 9 = 25 + 10 - 9 = 26
$$

✔ Answer: $26$

---

#### (c) $\lim_{x \to 1} \frac{3x^2 + 2x - 4}{x^2 + 5x - 4}$

Substitute $x = 1$:

Numerator: $3(1)^2 + 2(1) - 4 = 3 + 2 - 4 = 1$

Denominator: $(1)^2 + 5(1) - 4 = 1 + 5 - 4 = 2$

So,

$$
\frac{1}{2}
$$

✔ Answer: $\frac{1}{2}$

---

#### (d) $\lim_{x \to 3} \frac{6x^2 + 3x - 12}{2x^2 + x + 1}$

Substitute $x = 3$:

Numerator: $6(9) + 3(3) - 12 = 54 + 9 - 12 = 51$

Denominator: $2(9) + 3 + 1 = 18 + 3 + 1 = 22$

$$
\frac{51}{22}
$$

✔ Answer: $\frac{51}{22}$

---

2. Compute the following limits:

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

Factor numerator and denominator:

Numerator: $x(4x^2 - x + 2)$
Denominator: $x(3x + 4)$

Cancel $x$ (valid as $x \to 0$, $x \ne 0$):

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

✔ Answer: $\frac{1}{2}$

---

#### (b) $\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}$

Note: $x^3 - 64 = x^3 - 4^3 = (x - 4)(x^2 + 4x + 16)$
$x^2 - 16 = (x - 4)(x + 4)$

Cancel $(x - 4)$:

$$
\lim_{x \to 4} \frac{x^2 + 4x + 16}{x + 4} = \frac{16 + 16 + 16}{8} = \frac{48}{8} = 6
$$

✔ Answer: $6$

---

#### (c) $\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x - a}$

This is the definition of derivative of $f(x) = x^{2/3}$ at $x = a$, or use substitution.

Let $u = x^{1/3}, v = a^{1/3}$, then $x = u^3, a = v^3$, so:

$$
\frac{u^2 - v^2}{u^3 - v^3} = \frac{(u - v)(u + v)}{(u - v)(u^2 + uv + v^2)} = \frac{u + v}{u^2 + uv + v^2}
$$

As $x \to a$, $u \to v$, so:

$$
\frac{2v}{3v^2} = \frac{2}{3v} = \frac{2}{3a^{1/3}}
$$

Alternatively, use derivative rule: $f'(x) = \frac{2}{3}x^{-1/3}$, so limit is $\frac{2}{3}a^{-1/3}$

✔ Answer: $\frac{2}{3}a^{-1/3}$

---

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

Factor numerator: $x^2 + 3x - 4 = (x + 4)(x - 1)$

So:

$$
\frac{(x + 4)(x - 1)}{x - 1} = x + 4 \quad \text{for } x \ne 1
$$

Limit as $x \to 1$: $1 + 4 = 5$

✔ Answer: $5$

---

#### (e) $\lim_{x \to 2} \frac{x^2 - 5x + 6}{x^2 - x - 2}$

Factor both:

Numerator: $x^2 - 5x + 6 = (x - 2)(x - 3)$
Denominator: $x^2 - x - 2 = (x - 2)(x + 1)$

Cancel $(x - 2)$:

$$
\frac{x - 3}{x + 1} \to \frac{-1}{3} \quad \text{as } x \to 2
$$

✔ Answer: $-\frac{1}{3}$

---

#### (f) $\lim_{x \to 2} \frac{x^2 - 4x + 4}{x^2 - 7x + 10}$

Numerator: $(x - 2)^2$
Denominator: $x^2 - 7x + 10 = (x - 2)(x - 5)$

So:

$$
\frac{(x - 2)^2}{(x - 2)(x - 5)} = \frac{x - 2}{x - 5} \to \frac{0}{-3} = 0
$$

✔ Answer: $0$

---

#### (g) $\lim_{x \to a} \frac{\sqrt{3x} - \sqrt{2x + a}}{2(x - a)}$

Let’s rationalize the numerator.

Multiply numerator and denominator by conjugate:

$$
\frac{\sqrt{3x} - \sqrt{2x + a}}{2(x - a)} \cdot \frac{\sqrt{3x} + \sqrt{2x + a}}{\sqrt{3x} + \sqrt{2x + a}} = \frac{(3x) - (2x + a)}{2(x - a)(\sqrt{3x} + \sqrt{2x + a})}
$$

Simplify numerator: $3x - 2x - a = x - a$

So:

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

Now take limit as $x \to a$:

$$
\frac{1}{2(\sqrt{3a} + \sqrt{2a + a})} = \frac{1}{2(\sqrt{3a} + \sqrt{3a})} = \frac{1}{2(2\sqrt{3a})} = \frac{1}{4\sqrt{3a}}
$$

✔ Answer: $\frac{1}{4\sqrt{3a}}$

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#### (h) $\lim_{x \to a} \frac{\sqrt{2x} - \sqrt{3x - a}}{\sqrt{x} - \sqrt{a}}$

Let’s rationalize numerator and denominator.

First, multiply numerator and denominator by conjugate of denominator: $\sqrt{x} + \sqrt{a}$

But better to factor or use substitution. Let $x = a + h$, $h \to 0$. But let's try rationalizing.

Let’s write:

Let $A = \sqrt{2x} - \sqrt{3x - a}$, $B = \sqrt{x} - \sqrt{a}$

Multiply numerator and denominator by conjugate of numerator and denominator.

But perhaps easier to use Taylor expansion or substitution.

Let $x \to a$, so try direct substitution:

Numerator: $\sqrt{2a} - \sqrt{3a - a} = \sqrt{2a} - \sqrt{2a} = 0$

Denominator: $\sqrt{a} - \sqrt{a} = 0$

So indeterminate form $0/0$. Use L’Hôpital’s Rule.

Let:
$$
f(x) = \sqrt{2x} - \sqrt{3x - a}, \quad g(x) = \sqrt{x} - \sqrt{a}
$$

Then:
$$
f'(x) = \frac{2}{2\sqrt{2x}} - \frac{3}{2\sqrt{3x - a}} = \frac{1}{\sqrt{2x}} - \frac{3}{2\sqrt{3x - a}}
$$

Wait, derivative of $\sqrt{2x} = \frac{2}{2\sqrt{2x}} = \frac{1}{\sqrt{2x}}$

Derivative of $\sqrt{3x - a} = \frac{3}{2\sqrt{3x - a}}$

So:
$$
f'(x) = \frac{1}{\sqrt{2x}} - \frac{3}{2\sqrt{3x - a}}
$$

$g'(x) = \frac{1}{2\sqrt{x}}$

Now evaluate at $x = a$:

$f'(a) = \frac{1}{\sqrt{2a}} - \frac{3}{2\sqrt{2a}} = \left(1 - \frac{3}{2}\right)\frac{1}{\sqrt{2a}} = -\frac{1}{2\sqrt{2a}}$

$g'(a) = \frac{1}{2\sqrt{a}}$

So limit:
$$
\frac{f'(a)}{g'(a)} = \frac{-\frac{1}{2\sqrt{2a}}}{\frac{1}{2\sqrt{a}}} = -\frac{1}{2\sqrt{2a}} \cdot \frac{2\sqrt{a}}{1} = -\frac{\sqrt{a}}{\sqrt{2a}} = -\frac{1}{\sqrt{2}}
$$

✔ Answer: $-\frac{1}{\sqrt{2}}$

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#### (i) $\lim_{x \to 1} \frac{\sqrt{2x} - \sqrt{3 - x^2}}{x - 1}$

At $x = 1$: numerator: $\sqrt{2} - \sqrt{3 - 1} = \sqrt{2} - \sqrt{2} = 0$, denominator: 0 → $0/0$

Use L’Hôpital’s Rule.

Let $f(x) = \sqrt{2x} - \sqrt{3 - x^2}$, $g(x) = x - 1$

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

$g'(x) = 1$

So:
$$
\lim_{x \to 1} f'(x) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}
$$

✔ Answer: $\sqrt{2}$

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#### (j) $\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{6 - x^2}}{x - 2}$

Check $x = 2$: numerator: $\sqrt{2} - \sqrt{6 - 4} = \sqrt{2} - \sqrt{2} = 0$, denominator: 0 → $0/0$

Use L’Hôpital.

$f(x) = \sqrt{x} - \sqrt{6 - x^2}$, $g(x) = x - 2$

$f'(x) = \frac{1}{2\sqrt{x}} - \frac{-2x}{2\sqrt{6 - x^2}} = \frac{1}{2\sqrt{x}} + \frac{x}{\sqrt{6 - x^2}}$

At $x = 2$: $\frac{1}{2\sqrt{2}} + \frac{2}{\sqrt{6 - 4}} = \frac{1}{2\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{1}{2\sqrt{2}} + \frac{4}{2\sqrt{2}} = \frac{5}{2\sqrt{2}}$

So limit = $\frac{5}{2\sqrt{2}} = \frac{5\sqrt{2}}{4}$

✔ Answer: $\frac{5\sqrt{2}}{4}$

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#### (k) $\lim_{x \to 64} \frac{\sqrt[6]{x} - 2}{\sqrt[4]{x} - 4}$

Let $y = x^{1/12}$, then:

- $x = y^{12}$
- $\sqrt[6]{x} = x^{1/6} = y^2$
- $\sqrt[4]{x} = x^{1/4} = y^3$

So as $x \to 64$, $y^{12} = 64 = 2^6$, so $y = 2^{6/12} = 2^{1/2} = \sqrt{2}$? Wait — check:

$64 = 2^6$, so $x = 64 = 2^6$

We want $y^{12} = 2^6$, so $y = 2^{6/12} = 2^{1/2} = \sqrt{2}$

Then:
- $\sqrt[6]{x} = x^{1/6} = (2^6)^{1/6} = 2$
- $\sqrt[4]{x} = (2^6)^{1/4} = 2^{6/4} = 2^{3/2} = 2\sqrt{2}$

Wait! So denominator: $\sqrt[4]{x} - 4 = 2\sqrt{2} - 4 \approx 2.828 - 4 = -1.172$, but at $x=64$, numerator: $2 - 2 = 0$, denominator: $2\sqrt{2} - 4 \ne 0$?

Wait: $\sqrt[4]{64} = (64)^{1/4} = (2^6)^{1/4} = 2^{6/4} = 2^{1.5} = 2\sqrt{2} \approx 2.828$, but the expression says $\sqrt[4]{x} - 4$, which is not zero.

But numerator: $\sqrt[6]{64} = 64^{1/6} = (2^6)^{1/6} = 2$, so $2 - 2 = 0$

Denominator: $64^{1/4} - 4 = 2^{3/2} - 4 \approx 2.828 - 4 = -1.172 \ne 0$

So this is not indeterminate? But the limit is as $x \to 64$, and if denominator is not zero, we can plug in?

Wait: Is it possible that the problem has a typo?

Wait: Let’s recheck:

We have:

$\sqrt[6]{x} \to \sqrt[6]{64} = 64^{1/6} = (2^6)^{1/6} = 2$

$\sqrt[4]{x} \to 64^{1/4} = (2^6)^{1/4} = 2^{6/4} = 2^{3/2} = 2\sqrt{2} \approx 2.828$

So denominator: $2\sqrt{2} - 4 \ne 0$

So the limit is:

$$
\frac{2 - 2}{2\sqrt{2} - 4} = \frac{0}{2\sqrt{2} - 4} = 0
$$

But wait — is the denominator approaching zero? No, it's approaching $2\sqrt{2} - 4 \ne 0$, so yes, limit is 0.

But let’s double-check: maybe the expression was meant to be $\sqrt[4]{x} - \sqrt[4]{64}$ or something?

Wait: The expression is:

$$
\lim_{x \to 64} \frac{\sqrt[6]{x} - 2}{\sqrt[4]{x} - 4}
$$

At $x = 64$, numerator = 0, denominator = $64^{1/4} - 4 = 2^{3/2} - 4 = 2\sqrt{2} - 4 \ne 0$

So the limit is simply $0 / (2\sqrt{2} - 4) = 0$

✔ Answer: $0$

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#### (l) $\lim_{x \to a} \frac{\sqrt{3a - x} - \sqrt{x + a}}{4(x - a)}$

At $x = a$: numerator: $\sqrt{3a - a} - \sqrt{a + a} = \sqrt{2a} - \sqrt{2a} = 0$, denominator: 0 → $0/0$

Use L’Hôpital.

Let $f(x) = \sqrt{3a - x} - \sqrt{x + a}$, $g(x) = 4(x - a)$

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

$g'(x) = 4$

So limit:

$$
\frac{f'(a)}{g'(a)} = \frac{ \frac{-1}{2\sqrt{2a}} - \frac{1}{2\sqrt{2a}} }{4} = \frac{ -\frac{1}{\sqrt{2a}} }{4} = -\frac{1}{4\sqrt{2a}}
$$

✔ Answer: $-\frac{1}{4\sqrt{2a}}$

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3. Calculate the following limits:

These are limits at infinity.

#### (a) $\lim_{x \to \infty} \frac{2x^2}{3x^2 + 2}$

Divide numerator and denominator by $x^2$:

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

✔ Answer: $\frac{2}{3}$

---

#### (b) $\lim_{x \to \infty} \frac{3x^2 - 4}{4x^2}$

Divide by $x^2$: $\frac{3 - \frac{4}{x^2}}{4} \to \frac{3}{4}$

✔ Answer: $\frac{3}{4}$

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#### (c) $\lim_{x \to \infty} \frac{4x^2 + 3x + 2}{5x^2 + 4x - 3}$

Divide by $x^2$: $\frac{4 + \frac{3}{x} + \frac{2}{x^2}}{5 + \frac{4}{x} - \frac{3}{x^2}} \to \frac{4}{5}$

✔ Answer: $\frac{4}{5}$

---

#### (d) $\lim_{x \to \infty} \frac{5x^2 + 2x - 7}{3x^2 + 5x + 2}$

Divide by $x^2$: $\frac{5 + \frac{2}{x} - \frac{7}{x^2}}{3 + \frac{5}{x} + \frac{2}{x^2}} \to \frac{5}{3}$

✔ Answer: $\frac{5}{3}$

---

4. Calculate the following limits:

#### (a) $\lim_{x \to \infty} (\sqrt{x} - \sqrt{x - 3})$

Rationalize:

$$
\sqrt{x} - \sqrt{x - 3} = \frac{(\sqrt{x} - \sqrt{x - 3})(\sqrt{x} + \sqrt{x - 3})}{\sqrt{x} + \sqrt{x - 3}} = \frac{x - (x - 3)}{\sqrt{x} + \sqrt{x - 3}} = \frac{3}{\sqrt{x} + \sqrt{x - 3}}
$$

As $x \to \infty$, denominator $\to \infty$, so limit = 0

✔ Answer: $0$

---

#### (b) $\lim_{x \to \infty} (\sqrt{x - a} - \sqrt{x - b})$

Assume $a < b$. Then:

$$
\sqrt{x - a} - \sqrt{x - b} = \frac{(x - a) - (x - b)}{\sqrt{x - a} + \sqrt{x - b}} = \frac{b - a}{\sqrt{x - a} + \sqrt{x - b}} \to 0
$$

✔ Answer: $0$

---

#### (c) $\lim_{x \to \infty} (\sqrt{3x} - \sqrt{x - 3})$

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

Divide numerator and denominator by $\sqrt{x}$:

Numerator: $2\sqrt{x} + \frac{3}{\sqrt{x}}$

Denominator: $\sqrt{3} + \sqrt{1 - \frac{3}{x}} \to \sqrt{3} + 1$

So overall: $\frac{2\sqrt{x}}{\sqrt{3} + 1} \to \infty$

So limit is $\infty$

✔ Answer: $\infty$

---

#### (d) $\lim_{x \to \infty} \sqrt{x}(\sqrt{x} - \sqrt{x - a})$

$$
= \sqrt{x} \cdot \left( \frac{(\sqrt{x} - \sqrt{x - a})(\sqrt{x} + \sqrt{x - a})}{\sqrt{x} + \sqrt{x - a}} \right) = \sqrt{x} \cdot \left( \frac{a}{\sqrt{x} + \sqrt{x - a}} \right)
$$

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

✔ Answer: $\frac{a}{2}$

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5. (a) $\lim_{x \to 2} \frac{x - \sqrt{8 - x^2}}{\sqrt{x^2 + 12} - 4}$

At $x = 2$: numerator: $2 - \sqrt{8 - 4} = 2 - \sqrt{4} = 2 - 2 = 0$

Denominator: $\sqrt{4 + 12} - 4 = \sqrt{16} - 4 = 4 - 4 = 0$ → $0/0$

Use L’Hôpital.

Let $f(x) = x - \sqrt{8 - x^2}$, $g(x) = \sqrt{x^2 + 12} - 4$

$f'(x) = 1 - \frac{-2x}{2\sqrt{8 - x^2}} = 1 + \frac{x}{\sqrt{8 - x^2}}$

$g'(x) = \frac{2x}{2\sqrt{x^2 + 12}} = \frac{x}{\sqrt{x^2 + 12}}$

At $x = 2$:

$f'(2) = 1 + \frac{2}{\sqrt{8 - 4}} = 1 + \frac{2}{2} = 1 + 1 = 2$

$g'(2) = \frac{2}{\sqrt{4 + 12}} = \frac{2}{4} = \frac{1}{2}$

So limit = $2 / (1/2) = 4$

✔ Answer: $4$

---

#### (b) $\lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x}}$

At $x = 1$: numerator: $1 - \sqrt{2 - 1} = 1 - 1 = 0$

Denominator: $2 - \sqrt{2 + 2} = 2 - \sqrt{4} = 2 - 2 = 0$ → $0/0$

Use L’Hôpital.

$f(x) = x - \sqrt{2 - x^2}$, $f'(x) = 1 + \frac{x}{\sqrt{2 - x^2}}$

$g(x) = 2x - \sqrt{2 + 2x}$, $g'(x) = 2 - \frac{2}{2\sqrt{2 + 2x}} = 2 - \frac{1}{\sqrt{2 + 2x}}$

At $x = 1$:

$f'(1) = 1 + \frac{1}{\sqrt{1}} = 1 + 1 = 2$

$g'(1) = 2 - \frac{1}{\sqrt{4}} = 2 - \frac{1}{2} = \frac{3}{2}$

Limit = $2 / (3/2) = 4/3$

✔ Answer: $\frac{4}{3}$

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## ✔ Final Answers Summary:

1.

(a) $0$
(b) $26$
(c) $\frac{1}{2}$
(d) $\frac{51}{22}$

2.

(a) $\frac{1}{2}$
(b) $6$
(c) $\frac{2}{3}a^{-1/3}$
(d) $5$
(e) $-\frac{1}{3}$
(f) $0$
(g) $\frac{1}{4\sqrt{3a}}$
(h) $-\frac{1}{\sqrt{2}}$
(i) $\sqrt{2}$
(j) $\frac{5\sqrt{2}}{4}$
(k) $0$
(l) $-\frac{1}{4\sqrt{2a}}$

3.

(a) $\frac{2}{3}$
(b) $\frac{3}{4}$
(c) $\frac{4}{5}$
(d) $\frac{5}{3}$

4.

(a) $0$
(b) $0$
(c) $\infty$
(d) $\frac{a}{2}$

5.

(a) $4$
(b) $\frac{4}{3}$

Let me know if you'd like any explanation in more detail!