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

Let's solve each of these limits step by step. Since you've uploaded an image with multiple limit problems (a) through (z), I'll go through them one by one, explaining the method and providing the solution.

---

General Strategy:

- For limits as $ x \to \infty $ or $ x \to -\infty $:
- Polynomials: Compare degrees of numerator and denominator.
- If degree(num) < degree(denom): limit is 0.
- If equal: ratio of leading coefficients.
- If greater: limit is $ \pm\infty $ depending on signs.
- Rational expressions involving roots: Factor or divide numerator and denominator by highest power of $ x $.
- Exponential-type expressions: Use asymptotic behavior or logarithmic approximations if needed.
- Indeterminate forms like $ \frac{\infty}{\infty} $: Use L’Hôpital’s Rule or algebraic simplification.

We’ll proceed systematically.

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a) $ \lim_{x \to \infty} \frac{x^2 - 1}{2x^2 + 1} $

Divide numerator and denominator by $ x^2 $:

$$
= \lim_{x \to \infty} \frac{1 - \frac{1}{x^2}}{2 + \frac{1}{x^2}} = \frac{1 - 0}{2 + 0} = \boxed{\frac{1}{2}}
$$

---

b) $ \lim_{x \to -\infty} \frac{x^3 + x^2 - 4}{2x^3 + x + 11} $

Leading terms: $ x^3 / 2x^3 $. Degrees are same.

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

But check sign: As $ x \to -\infty $, both numerator and denominator → $ -\infty $, so ratio is positive.

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

---

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

Degree num = 2, den = 3 → higher degree in denominator ⇒ limit is:

$$
\boxed{0}
$$

---

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

Simplify inside:

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

Now take limit:

$$
\lim_{x \to \infty} \frac{-2x}{x^2 + 2} = \lim_{x \to \infty} \frac{-2/x}{1 + 2/x^2} = \frac{0}{1} = \boxed{0}
$$

---

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

Same degree: $ x^2 $. Ratio of leading coefficients: $ 1/3 $

$$
\boxed{\frac{1}{3}}
$$

---

f) $ \lim_{x \to \infty} \frac{x(x - 1)(x - 2)}{x^2 + 6x - 9} $

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

Denominator: $ x^2 + 6x - 9 $

So we have $ \frac{x^3 + \dots}{x^2 + \dots} $: degree num > den → goes to $ \infty $

As $ x \to \infty $, positive, so:

$$
\boxed{\infty}
$$

---

g) $ \lim_{x \to \infty} \frac{\sqrt{x^2 + 9}}{x + 3} $

Note: $ \sqrt{x^2 + 9} \sim |x| $ as $ x \to \infty $. Since $ x > 0 $, $ |x| = x $

So approximate: $ \frac{x}{x + 3} \to 1 $

But let's be precise:

Factor $ x $:

$$
= \frac{|x|\sqrt{1 + \frac{9}{x^2}}}{x + 3} = \frac{x \sqrt{1 + \frac{9}{x^2}}}{x(1 + \frac{3}{x})} = \frac{\sqrt{1 + \frac{9}{x^2}}}{1 + \frac{3}{x}} \to \frac{1}{1} = \boxed{1}
$$

---

h) $ \lim_{x \to \infty} \left( \frac{x^2 + x - 1}{2x^2 - x + 1} \right)^3 $

First compute inner limit:

$$
\frac{x^2 + x - 1}{2x^2 - x + 1} \to \frac{1}{2} \quad \text{(same degree)}
$$

Then cube it:

$$
\left(\frac{1}{2}\right)^3 = \boxed{\frac{1}{8}}
$$

---

i) $ \lim_{x \to \infty} \frac{x^2 + 2x + 1}{5x} $

Numerator: $ (x+1)^2 $, but degree 2, denominator degree 1 → $ \infty $

$$
\boxed{\infty}
$$

---

j) $ \lim_{x \to \infty} \frac{x^3 + x^4 - 1}{2x^5 + x - x^2} $

Rewrite: numerator: $ x^4 + x^3 - 1 $, degree 4
Denominator: $ 2x^5 - x^2 + x $, degree 5

So degree den > num ⇒ limit is:

$$
\boxed{0}
$$

---

k) $ \lim_{x \to \infty} \frac{(\sqrt{x^2 + 1} + x)^2}{\sqrt[3]{x^6 + 1}} $

Analyze numerator and denominator separately.

#### Numerator:
$ (\sqrt{x^2 + 1} + x)^2 $

Note: $ \sqrt{x^2 + 1} = x\sqrt{1 + \frac{1}{x^2}} \approx x(1 + \frac{1}{2x^2}) = x + \frac{1}{2x} $

So $ \sqrt{x^2 + 1} + x \approx x + \frac{1}{2x} + x = 2x + \frac{1}{2x} $

Square: $ (2x + \frac{1}{2x})^2 = 4x^2 + 2 + \frac{1}{4x^2} \to \infty $, dominated by $ 4x^2 $

More precisely:

Let’s rationalize or simplify:

Let $ A = \sqrt{x^2 + 1} + x $

Multiply conjugate: $ A \cdot (\sqrt{x^2 + 1} - x) = (x^2 + 1) - x^2 = 1 $

So $ A = \frac{1}{\sqrt{x^2 + 1} - x} $

But perhaps better to just expand:

$$
(\sqrt{x^2 + 1} + x)^2 = x^2 + 1 + 2x\sqrt{x^2 + 1} + x^2 = 2x^2 + 1 + 2x\sqrt{x^2 + 1}
$$

Now $ \sqrt{x^2 + 1} \sim x $, so $ 2x \cdot x = 2x^2 $, so total ~ $ 2x^2 + 1 + 2x^2 = 4x^2 $

So numerator ~ $ 4x^2 $

#### Denominator:
$ \sqrt[3]{x^6 + 1} = (x^6 + 1)^{1/3} \sim x^{6/3} = x^2 $

So overall:

$$
\frac{4x^2}{x^2} = 4 \Rightarrow \boxed{4}
$$

Wait — but let’s be more careful:

Numerator: $ (\sqrt{x^2 + 1} + x)^2 \sim (x + x)^2 = (2x)^2 = 4x^2 $

Denominator: $ (x^6 + 1)^{1/3} \sim x^2 $

So ratio → $ 4x^2 / x^2 = \boxed{4} $

Yes.

---

l) $ \lim_{x \to \infty} \frac{x^6 + 7x^4 - 40}{1 - x - 5x^7} $

Numerator: degree 6
Denominator: $ -5x^7 + \dots $ → degree 7

So degree den > num → limit is:

$$
\boxed{0}
$$

---

m) $ \lim_{x \to \infty} \frac{(x+1)(x-2)}{3x^2 + 6x - 5} $

Numerator: $ x^2 - x - 2 $

Denominator: $ 3x^2 + 6x - 5 $

Same degree: ratio of leading coefficients: $ 1/3 $

$$
\boxed{\frac{1}{3}}
$$

---

n) $ \lim_{x \to \infty} \frac{\sqrt{x^2 + 1}}{x} $

$$
= \frac{|x|\sqrt{1 + \frac{1}{x^2}}}{x} = \frac{x \cdot \sqrt{1 + \frac{1}{x^2}}}{x} = \sqrt{1 + \frac{1}{x^2}} \to \boxed{1}
$$

---

o) $ \lim_{x \to \infty} \left( \frac{3x^2 + 2x + 1}{x^2 - 3x + 2} \right)^4 $

Inner fraction: degrees same → $ \frac{3}{1} = 3 $

So $ 3^4 = \boxed{81} $

---

p) $ \lim_{x \to \infty} \frac{5x^3 - x^2 + x}{1 - x - 3x^2} $

Numerator: $ 5x^3 + \dots $
Denominator: $ -3x^2 + \dots $

Degree num > den → $ \infty $

But sign: leading term $ \frac{5x^3}{-3x^2} = -\frac{5}{3}x \to -\infty $

So:

$$
\boxed{-\infty}
$$

---

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

Numerator: $ -3x^3 + x + 1 $
Denominator: $ 3x^3 + x^2 + 1 $

Same degree: $ x^3 $. Leading coefficients: $ -3 / 3 = -1 $

$$
\boxed{-1}
$$

---

r) $ \lim_{x \to \infty} \left( \frac{x^3 - 8}{x^4 + 16} \right)^{10} $

Inside: $ \frac{x^3}{x^4} = \frac{1}{x} \to 0 $

So $ ( \to 0 )^{10} = \boxed{0} $

---

s) $ \lim_{x \to \infty} \frac{(x+3)(x+4)(x+5)}{x^4 + x - 11} $

Numerator: cubic → $ x^3 $
Denominator: quartic → $ x^4 $

So degree den > num → $ \boxed{0} $

---

t) $ \lim_{x \to -\infty} \frac{8x - 2x^5 + x^6}{11x + 5x^3 + 3x^5} $

Numerator: $ x^6 - 2x^5 + 8x $ → leading term $ x^6 $
Denominator: $ 3x^5 + 5x^3 + 11x $ → leading term $ 3x^5 $

So ratio: $ \frac{x^6}{3x^5} = \frac{1}{3}x \to -\infty $ because $ x \to -\infty $

So:

$$
\boxed{-\infty}
$$

---

u) $ \lim_{x \to \infty} \left( \frac{x^3}{2x^2 - 1} - \frac{x^2}{2x + 1} \right) $

Break into two parts:

First term: $ \frac{x^3}{2x^2 - 1} \sim \frac{x^3}{2x^2} = \frac{x}{2} $

Second term: $ \frac{x^2}{2x + 1} \sim \frac{x^2}{2x} = \frac{x}{2} $

So difference: $ \frac{x}{2} - \frac{x}{2} = 0 $? But need precision.

Compute:

Let’s write:

$$
\frac{x^3}{2x^2 - 1} = \frac{x}{2} \cdot \frac{1}{1 - \frac{1}{2x^2}} \approx \frac{x}{2}(1 + \frac{1}{2x^2}) = \frac{x}{2} + \frac{1}{4x}
$$

Similarly,

$$
\frac{x^2}{2x + 1} = \frac{x}{2} \cdot \frac{1}{1 + \frac{1}{2x}} \approx \frac{x}{2}(1 - \frac{1}{2x} + \cdots) = \frac{x}{2} - \frac{1}{4} + \cdots
$$

So difference:

$$
\left( \frac{x}{2} + \frac{1}{4x} \right) - \left( \frac{x}{2} - \frac{1}{4} \right) = \frac{1}{4x} + \frac{1}{4} \to \boxed{\frac{1}{4}}
$$

Wait — that suggests limit is $ \frac{1}{4} $, not zero.

Let’s do exact subtraction:

Let $ A = \frac{x^3}{2x^2 - 1} - \frac{x^2}{2x + 1} $

Get common denominator? Too messy.

Instead, factor out $ x $:

Write:

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

Now look at:

$$
\frac{x^2}{2x^2 - 1} \to \frac{1}{2}, \quad \frac{x}{2x + 1} \to \frac{1}{2}
$$

So $ \frac{1}{2} - \frac{1}{2} = 0 $, but multiplied by $ x $ → indeterminate form $ \infty \cdot 0 $

Need expansion.

Let’s use asymptotics:

- $ \frac{x^2}{2x^2 - 1} = \frac{1}{2} \cdot \frac{1}{1 - \frac{1}{2x^2}} \approx \frac{1}{2}(1 + \frac{1}{2x^2}) = \frac{1}{2} + \frac{1}{4x^2} $
- $ \frac{x}{2x + 1} = \frac{1}{2} \cdot \frac{1}{1 + \frac{1}{2x}} \approx \frac{1}{2}(1 - \frac{1}{2x} + \frac{1}{4x^2}) = \frac{1}{2} - \frac{1}{4x} + \frac{1}{8x^2} $

So difference:

$$
\left( \frac{1}{2} + \frac{1}{4x^2} \right) - \left( \frac{1}{2} - \frac{1}{4x} + \frac{1}{8x^2} \right) = \frac{1}{4x} + \left( \frac{1}{4} - \frac{1}{8} \right)\frac{1}{x^2} = \frac{1}{4x} + \frac{1}{8x^2}
$$

Now multiply by $ x $:

$$
x \cdot \left( \frac{1}{4x} + \frac{1}{8x^2} \right) = \frac{1}{4} + \frac{1}{8x} \to \boxed{\frac{1}{4}}
$$

✔ So final answer: $ \boxed{\frac{1}{4}} $

---

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

Simplify:

Let $ A = x^2 - \frac{x^4 - 1}{x^2 - 2} $

Write $ x^2 = \frac{x^2(x^2 - 2)}{x^2 - 2} = \frac{x^4 - 2x^2}{x^2 - 2} $

So:

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

Now:

$$
\lim_{x \to \infty} \frac{-2x^2 + 1}{x^2 - 2} = \frac{-2}{1} = \boxed{-2}
$$

---

w) $ \lim_{x \to \infty} \frac{(x-1)^{100}(6x+1)^{200}}{(3x+5)^{300}} $

All terms are polynomials. Let’s analyze degrees:

- $ (x-1)^{100} \sim x^{100} $
- $ (6x+1)^{200} \sim (6x)^{200} = 6^{200} x^{200} $
- $ (3x+5)^{300} \sim (3x)^{300} = 3^{300} x^{300} $

So total numerator: $ x^{100 + 200} = x^{300} $, denominator: $ x^{300} $

So ratio: $ \frac{6^{200} x^{300}}{3^{300} x^{300}} = \left( \frac{6^{200}}{3^{300}} \right) $

But $ 6^{200} = (2 \cdot 3)^{200} = 2^{200} \cdot 3^{200} $

So:

$$
\frac{2^{200} \cdot 3^{200}}{3^{300}} = 2^{200} \cdot 3^{-100} = \frac{2^{200}}{3^{100}} = \left( \frac{2^2}{3} \right)^{100} = \left( \frac{4}{3} \right)^{100}
$$

Wait — no: $ 2^{200} / 3^{100} = (2^2)^{100} / 3^{100} = (4/3)^{100} $

So:

$$
\boxed{\left( \frac{4}{3} \right)^{100}}
$$

But this is a constant, so limit is finite:

$$
\boxed{\left( \frac{4}{3} \right)^{100}}
$$

Wait — actually, we already factored all powers. The $ x^{300} $ cancels, so yes, limit is:

$$
\boxed{\left( \frac{4}{3} \right)^{100}}
$$

---

x) $ \lim_{x \to \infty} \frac{\sqrt[4]{x^5} + \sqrt[3]{x^2} + \sqrt[6]{x^8}}{\sqrt[3]{x^4 + 2}} $

Convert to exponents:

- $ x^{5/4} $
- $ x^{2/3} $
- $ x^{8/6} = x^{4/3} $
- Denominator: $ (x^4 + 2)^{1/3} \sim x^{4/3} $

Now compare growth:

- $ x^{5/4} = x^{1.25} $
- $ x^{4/3} \approx x^{1.333} $
- $ x^{2/3} \approx x^{0.666} $

So dominant term in numerator: $ x^{4/3} $, denominator: $ x^{4/3} $

So ratio → ratio of leading coefficients.

But all are monomials, so:

Numerator: $ x^{4/3} + x^{5/4} + x^{2/3} \sim x^{4/3} $

Denominator: $ x^{4/3} $

So ratio → $ \boxed{1} $

---

y) $ \lim_{x \to \infty} \frac{x^2(2x+1)(3x-2)}{2x^2(5x-8)(x+6)} $

Expand numerator and denominator.

Numerator: $ x^2(2x+1)(3x-2) $

First: $ (2x+1)(3x-2) = 6x^2 - 4x + 3x - 2 = 6x^2 - x - 2 $

Then $ x^2(6x^2 - x - 2) = 6x^4 - x^3 - 2x^2 $

Denominator: $ 2x^2(5x-8)(x+6) $

First: $ (5x-8)(x+6) = 5x^2 + 30x - 8x - 48 = 5x^2 + 22x - 48 $

Then $ 2x^2(5x^2 + 22x - 48) = 10x^4 + 44x^3 - 96x^2 $

So:

$$
\frac{6x^4 - x^3 - 2x^2}{10x^4 + 44x^3 - 96x^2} \to \frac{6}{10} = \boxed{\frac{3}{5}}
$$

---

z) $ \lim_{x \to \infty} \left( \frac{2x^8 + 8x^6 + 6x^4}{4x^8 - x^6 + 12x^4} \right)^5 $

Inside: degrees same (8). Leading terms: $ 2x^8 / 4x^8 = 1/2 $

So $ (1/2)^5 = \boxed{\frac{1}{32}} $

---

Z) $ \lim_{x \to \infty} \frac{(2x-3)^{20}(3x+2)^{30}}{(2x+1)^{50}} $

Let’s factor out highest powers.

Write:

- $ (2x - 3)^{20} = (2x)^{20} \left(1 - \frac{3}{2x}\right)^{20} $
- $ (3x + 2)^{30} = (3x)^{30} \left(1 + \frac{2}{3x}\right)^{30} $
- $ (2x + 1)^{50} = (2x)^{50} \left(1 + \frac{1}{2x}\right)^{50} $

So overall:

$$
\frac{(2x)^{20} (3x)^{30} \cdot \text{(terms → 1)}}{(2x)^{50} \cdot \text{(terms → 1)}} = \frac{(2^{20})(3^{30}) x^{20+30}}{2^{50} x^{50}} = \frac{3^{30}}{2^{30}} \cdot \frac{1}{x^{0}} = \left( \frac{3}{2} \right)^{30}
$$

Wait: $ x^{50} / x^{50} = 1 $

So:

$$
= \frac{2^{20} \cdot 3^{30}}{2^{50}} = 3^{30} \cdot 2^{-30} = \left( \frac{3}{2} \right)^{30}
$$

And the other factors → 1 as $ x \to \infty $

So limit is:

$$
\boxed{\left( \frac{3}{2} \right)^{30}}
$$

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| a) | $ \boxed{\frac{1}{2}} $ |
| b) | $ \boxed{\frac{1}{2}} $ |
| c) | $ \boxed{0} $ |
| d) | $ \boxed{0} $ |
| e) | $ \boxed{\frac{1}{3}} $ |
| f) | $ \boxed{\infty} $ |
| g) | $ \boxed{1} $ |
| h) | $ \boxed{\frac{1}{8}} $ |
| i) | $ \boxed{\infty} $ |
| j) | $ \boxed{0} $ |
| k) | $ \boxed{4} $ |
| l) | $ \boxed{0} $ |
| m) | $ \boxed{\frac{1}{3}} $ |
| n) | $ \boxed{1} $ |
| o) | $ \boxed{81} $ |
| p) | $ \boxed{-\infty} $ |
| q) | $ \boxed{-1} $ |
| r) | $ \boxed{0} $ |
| s) | $ \boxed{0} $ |
| t) | $ \boxed{-\infty} $ |
| u) | $ \boxed{\frac{1}{4}} $ |
| v) | $ \boxed{-2} $ |
| w) | $ \boxed{\left( \frac{4}{3} \right)^{100}} $ |
| x) | $ \boxed{1} $ |
| y) | $ \boxed{\frac{3}{5}} $ |
| z) | $ \boxed{\frac{1}{32}} $ |
| Z) | $ \boxed{\left( \frac{3}{2} \right)^{30}} $ |

Let me know if you want any explanation expanded!