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

We are given a list of 26 limit problems (a through Z), all of the form:

\[
\lim_{x \to \infty} \text{(rational expression or expression involving powers of } x\text{)}
\]

The key idea for limits at infinity of rational functions (ratios of polynomials) is to compare the degrees (highest powers) of numerator and denominator:

- If degree(numerator) < degree(denominator) → limit = 0
- If degree(numerator) = degree(denominator) → limit = ratio of leading coefficients
- If degree(numerator) > degree(denominator) → limit = ±∞ (sign depends on leading coefficients)

Also, for expressions like \(\sqrt{x^2 + a}\), as \(x \to \infty\), \(\sqrt{x^2 + a} \sim |x| = x\) (since \(x > 0\) when approaching \(+\infty\)).

Let’s compute each one carefully.

---

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

Degrees: both 2. Leading coefficients: 1 (num), 2 (den).
→ Limit = \(1/2\)

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

Degrees: both 3. Leading coeffs: 1 / 2 → limit = \(1/2\)

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

Num deg = 2, den deg = 3 → limit = 0

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

Simplify:
\[
\frac{x^3}{x^2 + 2} - x = \frac{x^3 - x(x^2 + 2)}{x^2 + 2} = \frac{x^3 - x^3 - 2x}{x^2 + 2} = \frac{-2x}{x^2 + 2}
\]
Now degrees: num = 1, den = 2 → limit = 0

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

Deg 2/2 → leading coeff ratio: \(1/3\)

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

Numerator expands to cubic: \(x^3 - 3x^2 + 2x\) → deg 3
Denominator deg 2 → limit = ∞ (positive, since leading coeff 1/1 > 0)
But careful: as \(x \to \infty\), sign is positive → limit = \(+\infty\)

However, in many standard calculus contexts, if not asked for sign, we say “does not exist” or “∞”. But since all others are finite or zero, maybe they expect ∞.

Let’s keep track — but maybe later we’ll see pattern.

Wait — the problem likely expects *finite* answers only? No, some may be infinite.

But let’s continue; maybe we’re supposed to evaluate all.

Hold on — perhaps the task is to match each limit with its value? But user didn’t specify what to do. Wait — the instruction says: “Solve the problem accurately.” And the image shows 26 limits labeled a–Z.

Since no further instruction, and typical homework like this asks: *Evaluate each limit.* So we need to give each value.

Let me compute all 26.

I’ll go systematically.

---

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

b) \(\frac{x^3 + x^2 - 4}{2x^3 + x + 11} \to \frac{1}{2}\)

c) \(\frac{3x^2 + 2x - 1}{x^3 - x + 2} \to 0\)

d) \(\frac{x^3}{x^2 + 2} - x = \frac{-2x}{x^2 + 2} \to 0\)

e) \(\frac{x^2 + 3x - 4}{3x^2 - 2x + 5} \to \frac{1}{3}\)

f) \(\frac{x(x-1)(x-2)}{x^2 + 6x - 9} = \frac{x^3 - 3x^2 + 2x}{x^2 + 6x - 9}\)
Divide numerator and denominator by \(x^2\):
= \(\frac{x - 3 + 2/x}{1 + 6/x - 9/x^2} \to \infty\) (since numerator → ∞, denominator → 1)
So limit = \(+\infty\)

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

Note: \(\sqrt{x^2 + 9} = x\sqrt{1 + 9/x^2} \sim x(1 + \frac{9}{2x^2}) = x + \frac{9}{2x} + \dots\)

So expression ≈ \(\frac{x}{x+3} \to 1\)

More rigor: divide numerator and denominator by \(x\) (x > 0):

\[
\frac{\sqrt{x^2 + 9}}{x + 3} = \frac{x\sqrt{1 + 9/x^2}}{x(1 + 3/x)} = \frac{\sqrt{1 + 9/x^2}}{1 + 3/x} \to \frac{1}{1} = 1
\]

→ limit = 1

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

Inside: ratio of quadratics → limit = \(1/2\), then cube: \((1/2)^3 = 1/8\)

i) \(\displaystyle \lim_{x \to \infty} \frac{x^2 + 2x + 1}{5x} = \frac{x^2(1 + 2/x + 1/x^2)}{5x} = \frac{x(1 + o(1))}{5} \to \infty\)

So \(+\infty\)

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

Rewrite numerator: \(x^4 + x^3 - 1\) → deg 4
Denominator: \(2x^5 - x^2 + x\) → deg 5
So deg num < deg den → limit = 0

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

First simplify numerator:
\[
(\sqrt{x^2 + 1} + x)^2 = x^2 + 1 + 2x\sqrt{x^2 + 1} + x^2 = 2x^2 + 1 + 2x\sqrt{x^2 + 1}
\]

As \(x \to \infty\), \(\sqrt{x^2 + 1} \sim x + \frac{1}{2x}\), so:
\(2x\sqrt{x^2 + 1} \sim 2x(x + \frac{1}{2x}) = 2x^2 + 1\)

So numerator ~ \(2x^2 + 1 + 2x^2 + 1 = 4x^2 + 2 \sim 4x^2\)

Denominator: \(3\sqrt{x^6 + 1} = 3x^3\sqrt{1 + 1/x^6} \sim 3x^3\)

So overall ~ \(\frac{4x^2}{3x^3} = \frac{4}{3x} \to 0\)

→ limit = 0

Alternative exact method: factor highest powers.

Numerator: factor \(x^2\):
\[
(\sqrt{x^2(1 + 1/x^2)} + x)^2 = (x\sqrt{1 + 1/x^2} + x)^2 = x^2(\sqrt{1 + 1/x^2} + 1)^2
\]

Denominator: \(3\sqrt{x^6(1 + 1/x^6)} = 3x^3\sqrt{1 + 1/x^6}\)

So expression =
\[
\frac{x^2(\sqrt{1 + 1/x^2} + 1)^2}{3x^3\sqrt{1 + 1/x^6}} = \frac{(\sqrt{1 + 1/x^2} + 1)^2}{3x\sqrt{1 + 1/x^6}} \to \frac{(1 + 1)^2}{3x \cdot 1} = \frac{4}{3x} \to 0
\]

Yes, 0.

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

Num deg = 6, den deg = 7 → limit = 0

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

Numerator: \(x^2 - x - 2\) → deg 2
Denominator: \(3x^2 + 6x -5\) → deg 2
Leading coeff ratio: \(1/3\)

→ limit = \(1/3\)

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

= \(\frac{x\sqrt{1 + 1/x^2}}{x} = \sqrt{1 + 1/x^2} \to 1\)

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

Inside → \(3/1 = 3\), then ^4 → \(3^4 = 81\)

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

Num deg = 3, den deg = 2 → limit = ±∞
Leading terms: \(5x^3 / (-3x^2) = -\frac{5}{3}x \to -\infty\)

So limit = \(-\infty\)

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

Both deg 3. Leading coeffs: num = \(-3\), den = \(3\) → limit = \(-1\)

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

Inside: deg num = 3, den = 4 → fraction → 0
Then 0^10 = 0

→ limit = 0

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

Numerator: cubic (degree 3), denominator: degree 4 → limit = 0

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

Reorder: numerator = \(x^6 - 2x^5 + 8x\) → deg 6
Denominator = \(3x^5 + 5x^3 + 11x\) → deg 5
So deg num > deg den → limit = ±∞
Leading term: \(x^6 / 3x^5 = \frac{1}{3}x \to +\infty\)

→ \(+\infty\)

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

Compute each term:

First term: \(\frac{x^3}{2x^2 - 1} = \frac{x}{2 - 1/x^2} = \frac{x}{2}(1 + \frac{1}{2x^2} + \dots) \approx \frac{x}{2} + \frac{1}{4x} + \dots\)

Second term: \(\frac{x^2}{2x + 1} = \frac{x}{2 + 1/x} = \frac{x}{2}(1 - \frac{1}{2x} + \dots) = \frac{x}{2} - \frac{1}{4} + \dots\)

Subtract:
\(\left(\frac{x}{2} + \frac{1}{4x} + \dots\right) - \left(\frac{x}{2} - \frac{1}{4} + \dots\right) = \frac{1}{4} + \frac{1}{4x} + \dots \to \frac{1}{4}\)

Let’s do exact algebra:

Let’s combine over common denominator:

\[
\frac{x^3}{2x^2 - 1} - \frac{x^2}{2x + 1} = \frac{x^3(2x + 1) - x^2(2x^2 - 1)}{(2x^2 - 1)(2x + 1)}
\]

Numerator:
\(x^3(2x + 1) = 2x^4 + x^3\)
\(x^2(2x^2 - 1) = 2x^4 - x^2\)
Subtract: \((2x^4 + x^3) - (2x^4 - x^2) = x^3 + x^2\)

Denominator: \((2x^2 - 1)(2x + 1) = 4x^3 + 2x^2 - 2x - 1\)

So expression = \(\frac{x^3 + x^2}{4x^3 + 2x^2 - 2x - 1}\)

Divide numerator & denominator by \(x^3\):

= \(\frac{1 + 1/x}{4 + 2/x - 2/x^2 - 1/x^3} \to \frac{1}{4}\)

→ limit = \(1/4\)

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

Simplify inside:

\[
x^2 - \frac{x^4 - 1}{x^2 - 2} = \frac{x^2(x^2 - 2) - (x^4 - 1)}{x^2 - 2} = \frac{x^4 - 2x^2 - x^4 + 1}{x^2 - 2} = \frac{-2x^2 + 1}{x^2 - 2}
\]

Now limit = \(\frac{-2x^2 + 1}{x^2 - 2} \to -2\)

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

We can use leading behavior:

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

So overall ~ \(\frac{x^{100} \cdot 6^{200} x^{200}}{3^{300} x^{300}} = \frac{6^{200}}{3^{300}} x^{300 - 300} = \frac{6^{200}}{3^{300}}\)

Note: \(6 = 2 \cdot 3\), so \(6^{200} = 2^{200} \cdot 3^{200}\)

Thus:
\[
\frac{2^{200} \cdot 3^{200}}{3^{300}} = 2^{200} \cdot 3^{-100} = \left(\frac{2^2}{3}\right)^{100} = \left(\frac{4}{3}\right)^{100}
\]

Wait: \(2^{200} / 3^{100} = (2^2)^{100} / 3^{100} = (4/3)^{100}\)

Yes.

So limit = \(\left(\frac{4}{3}\right)^{100}\)

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

Interpret roots:

\(\sqrt[4]{x^5} = x^{5/4}\)
\(\sqrt[5]{x^3} = x^{3/5}\)
\(\sqrt[6]{x^8} = x^{8/6} = x^{4/3}\)
Denominator: \(\sqrt[3]{x^4} = x^{4/3}\), plus 2.

So numerator dominant term: compare exponents:
- \(5/4 = 1.25\)
- \(3/5 = 0.6\)
- \(4/3 ≈ 1.333...\)

So largest exponent is \(4/3\), same as denominator’s leading term.

So write:

Numerator = \(x^{4/3} + x^{5/4} + x^{3/5}\)
But note: \(x^{5/4} = x^{1.25} < x^{4/3} ≈ x^{1.333}\), so indeed \(x^{4/3}\) dominates.

So factor \(x^{4/3}\):

Numerator = \(x^{4/3} \left(1 + x^{5/4 - 4/3} + x^{3/5 - 4/3}\right)\)

Compute exponents:
\(5/4 - 4/3 = \frac{15 - 16}{12} = -1/12\)
\(3/5 - 4/3 = \frac{9 - 20}{15} = -11/15\)

So numerator = \(x^{4/3}(1 + x^{-1/12} + x^{-11/15})\)

Denominator: \(x^{4/3} + 2 = x^{4/3}(1 + 2x^{-4/3})\)

So ratio:
\[
\frac{x^{4/3}(1 + x^{-1/12} + x^{-11/15})}{x^{4/3}(1 + 2x^{-4/3})} = \frac{1 + x^{-1/12} + x^{-11/15}}{1 + 2x^{-4/3}} \to \frac{1 + 0 + 0}{1 + 0} = 1
\]

→ limit = 1

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

Cancel \(x^2\) top and bottom:

= \(\frac{(2x + 1)(3x - 2)}{2(5x - 8)(x + 6)}\)

Expand leading terms:
Numerator: \(2x \cdot 3x = 6x^2\)
Denominator: \(2 \cdot 5x \cdot x = 10x^2\)

So limit = \(6/10 = 3/5\)

Check: more precisely,
\[
\frac{(2x+1)(3x-2)}{2(5x-8)(x+6)} = \frac{6x^2 - x - 2}{2(5x^2 - 2x - 48)} = \frac{6x^2 - x - 2}{10x^2 - 4x - 96}
\to \frac{6}{10} = \frac{3}{5}
\]

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

Leading behavior:
\((2x - 3)^{20} \sim (2x)^{20} = 2^{20} x^{20}\)
\((3x + 2)^{30} \sim (3x)^{30} = 3^{30} x^{30}\)
\((2x + 1)^{50} \sim (2x)^{50} = 2^{50} x^{50}\)

Multiply numerator: \(2^{20} \cdot 3^{30} \cdot x^{50}\)
Denominator: \(2^{50} x^{50}\)

Cancel \(x^{50}\): get \(\frac{2^{20} \cdot 3^{30}}{2^{50}} = 3^{30} / 2^{30} = \left(\frac{3}{2}\right)^{30}\)

→ limit = \(\left(\frac{3}{2}\right)^{30}\)

Now compile all answers:

a) 1/2
b) 1/2
c) 0
d) 0
e) 1/3
f) ∞
g) 1
h) 1/8
i) ∞
j) 0
k) 0
l) 0
m) 1/3
n) 1
o) 81
p) −∞
q) −1
r) 0
s) 0
t) ∞
u) 1/4
v) −2
w) (4/3)^100
x) 1
y) 3/5
z) (3/2)^30

But the problem may expect only finite values? However, the instruction says “Solve the problem accurately”, and there are infinite limits.

Wait — maybe the original task is multiple choice or matching? But user didn’t specify.

Given the format (a–Z), and no further instruction, safest is to list each limit value.

But the final answer section expects one answer? That suggests maybe the problem is to find a particular one? Hmm.

Re-read user message: “You are an educational assistant helping a student solve homework problems. The user uploaded an image containing a task. Solve the problem accurately.”

Possibility: In many textbooks, such a sheet asks: *Evaluate each of the following limits.* So the expected output is a list.

But the response format says: “Final Answer: [Provide only the final correct answer]” — singular.

That implies maybe only one is requested? But all 26 are shown.

Wait — maybe the actual task is to find which ones equal 0? Or something else? No clue.

Alternatively, perhaps the student is supposed to identify which limits are 0 — but not stated.

Given ambiguity, and instruction says “Solve the problem accurately”, and we have 26 independent limits, I think the proper thing is to provide all 26 results.

But the format requires a single Final Answer. Maybe the system expects the answer for a specific letter? No indication.

Let me double-check: The user set:
- ExplainAnswer: false
- TeachTopic: false
- GeneratePractice: false

So they just want the answer, no explanation — but we are told to do reasoning first, then final answer.

Given that, and that there are 26 items, perhaps the expected final answer is a list.

I will output the 26 values in order, separated by commas, as the final answer.

To be safe, I’ll format as:

a: 1/2, b: 1/2, c: 0, d: 0, e: 1/3, f: ∞, g: 1, h: 1/8, i: ∞, j: 0, k: 0, l: 0, m: 1/3, n: 1, o: 81, p: −∞, q: −1, r: 0, s: 0, t: ∞, u: 1/4, v: −2, w: (4/3)^100, x: 1, y: 3/5, z: (3/2)^30

But “∞” may be written as "infinity". In math contests, sometimes they accept “does not exist”, but here clearly diverges to ±∞.

Given the context, I'll use ∞ and −∞.

Now, before final, let me double-check a few tricky ones:

- u: we got 1/4 — confirmed via algebra.
- v: got −2 — yes, after simplification.
- w: (4/3)^100 — yes.
- z: (3/2)^30 — yes.
- x: 1 - yes, dominant powers cancel.
- p: -∞ — leading term 5x^3 / (-3x^2) = -(5/3)x → -∞ ✓
- f, i, t → +∞ ✓

All good.

Final Answer will be the list.

But the instruction says “Provide only the final correct answer” — so I will give the list as the final answer.

Since it's long, but required.

Let me write compactly:

Final Answer:
a) 1/2, b) 1/2, c) 0, d) 0, e) 1/3, f) ∞, g) 1, h) 1/8, i) ∞, j) 0, k) 0, l) 0, m) 1/3, n) 1, o) 81, p) −∞, q) −1, r) 0, s) 0, t) ∞, u) 1/4, v) −2, w) (4/3)^100, x) 1, y) 3/5, z) (3/2)^30