Comprehensive set of calculus limit problems for advanced algebra and pre-calculus students.

Let’s solve each limit problem one by one. We’re dealing with limits as x approaches infinity (∞). The key idea is: when x gets very large, the highest power of x in the numerator and denominator dominates — lower powers become negligible.

We’ll go letter by letter.

---

a)

limₓ→∞ (x² - 1)/(2x² + 1)
Divide numerator and denominator by x²:
= lim (1 - 1/x²)/(2 + 1/x²) → (1 - 0)/(2 + 0) = 1/2

---

b)

limₓ→∞ (x³ + x² - 4)/(2x³ + x + 11)
Divide by x³:
= (1 + 1/x - 4/x³)/(2 + 1/x² + 11/x³) → 1/2

---

c)

limₓ→∞ (3x² + 2x - 1)/(x³ - x + 2)
Numerator degree 2, denominator degree 3 → denominator grows faster → limit = 0

---

d)

limₓ→∞ [x³/(x² + 2) - x]
Combine terms:
= [x³ - x(x² + 2)] / (x² + 2) = [x³ - x³ - 2x]/(x² + 2) = (-2x)/(x² + 2)
Divide by x²: (-2/x)/(1 + 2/x²) → 0/-1? Wait — actually divide numerator and denominator by x:
= (-2)/(x + 2/x) → as x→∞, this goes to 0

Wait — let me recalculate:

Original: x³/(x²+2) - x = [x³ - x(x²+2)]/(x²+2) = [x³ - x³ - 2x]/(x²+2) = -2x/(x²+2)

Now divide numerator and denominator by x:
= -2 / (x + 2/x) → as x→∞, denominator → ∞ → whole thing → 0

---

e)

limₓ→∞ (x² + 3x - 4)/(3x² - 2x + 5)
Divide by x²: (1 + 3/x - 4/x²)/(3 - 2/x + 5/x²) → 1/3

---

f)

limₓ→∞ [x(x-1)(x-2)] / (x² + 6x - 9)
First expand numerator: x(x² - 3x + 2) = x³ - 3x² + 2x
Denominator: x² + 6x - 9
Degree num=3, den=2 → numerator grows faster → limit = ∞? But wait — we need to check sign.

As x→∞, leading term: x³ / x² = x → ∞ → so limit is ∞

But sometimes problems expect “does not exist” or just say infinity. Since it's going to positive infinity, we can write ∞

---

g)

limₓ→∞ √(x² + 9)/(x + 3)
For large x, √(x² + 9) ≈ |x| = x (since x→∞)
So ≈ x/(x+3) → divide by x: 1/(1 + 3/x) → 1/1 = 1

More precisely: factor x out of sqrt:
√(x²(1 + 9/x²)) = x√(1 + 9/x²)
So expression becomes: x√(1 + 9/x²) / (x + 3) = √(1 + 9/x²) / (1 + 3/x) → 1/1 = 1

---

h)

limₓ→∞ [(x² + x - 1)/(2x² - x + 1)]³
Inside: divide num/den by x² → (1 + 1/x - 1/x²)/(2 - 1/x + 1/x²) → 1/2
Then cube it: (1/2)³ = 1/8

---

i)

limₓ→∞ (x² + 2x + 1)/(5x)
Numerator degree 2, denominator degree 1 → grows like x²/x = x → ∞
So limit = ∞

---

j)

limₓ→∞ (x³ + x⁴ - 1)/(2x⁵ + x - x²)
Highest power in num: x⁴, den: x⁵ → den grows faster → limit = 0

---

k)

limₓ→∞ (√(x² + 1) + x)² / ∛(x + 1)
First, note: √(x² + 1) ≈ x for large x → so numerator ≈ (x + x)² = (2x)² = 4x²
Denominator: ∛(x⁶) = x²
So overall ≈ 4x² / x² = 4

Let’s be precise:

Numerator: [√(x²+1) + x]² = [x√(1+1/x²) + x]² = x²[√(1+1/x²) + 1]²
As x→∞, √(1+1/x²) → 1 → so inside → (1+1)² = 4 → numerator ~ 4x²

Denominator: ∛(x + 1) = x² ∛(1 + 1/x⁶) → x² * 1 = x²

So ratio → 4x² / x² = 4

---

l)

limₓ→∞ (x⁶ + 7x⁴ - 40)/(1 - x - 5x⁷)
Num degree 6, den degree 7 → den grows faster → limit = 0

Note: denominator has -5x⁷ → negative, but since we’re taking limit to ∞, and degree higher, still 0.

---

m)

limₓ→∞ [(x+1)(x-2)] / (3x² + 6x - 5)
Expand numerator: x² - x - 2
Denominator: 3x² + 6x - 5
Divide by x²: (1 - 1/x - 2/x²)/(3 + 6/x - 5/x²) → 1/3

---

n)

limₓ→∞ √(x² + 1)/x
= √(x²(1 + 1/x²))/x = |x|√(1 + 1/x²)/x
Since x→∞, |x|=x → so = x√(...)/x = √(1 + 1/x²) → √1 = 1

---

o)

limₓ→∞ [(3x² + 2x + 1)/(x² - 3x + 2)]⁴
Inside: divide by x² → (3 + 2/x + 1/x²)/(1 - 3/x + 2/x²) → 3/1 = 3
Then 3⁴ = 81

---

p)

limₓ→∞ (5x³ - x² + x)/(1 - x - 3x²)
Num degree 3, den degree 2 → num grows faster → limit = ±∞?

Leading terms: 5x³ / (-3x²) = -5/3 x → as x→∞, this → -∞

Because coefficient of x³ is positive, denominator leading term is -3x² → so overall negative and magnitude → ∞ → -∞

---

q)

limₓ→∞ (1 + x - 3x³)/(1 + x² + 3x³)
Divide by x³: (1/x³ + 1/x² - 3)/(1/x³ + 1/x + 3) → (-3)/3 = -1

---

r)

limₓ→∞ [(x³ - 8)/(x⁴ + 16)]¹⁰
Inside: num degree 3, den degree 4 → inside → 0
Then 0¹⁰ = 0

---

s)

limₓ→∞ [(x+3)(x+4)(x+5)] / (x⁴ + x - 11)
Numerator: cubic (degree 3), denominator degree 4 → limit = 0

---

t)

limₓ→∞ (8x - 2x⁵ + x⁶)/(11x + 5x³ + 3x⁵)
Highest power: num x⁶, den x⁵ → num grows faster → limit = ∞

Leading terms: x⁶ / 3x⁵ = x/3 → ∞ → ∞

---

u)

limₓ→∞ [x³/(2x² - 1) - x²/(2x + 1)]
Get common denominator or simplify each term.

First term: x³/(2x² - 1) ≈ x³/(2x²) = x/2
Second term: x²/(2x + 1) ≈ x²/(2x) = x/2
So difference ≈ x/2 - x/2 = 0? Need exact.

Compute:

= [x³(2x + 1) - x²(2x² - 1)] / [(2x² - 1)(2x + 1)]

Numerator: x³(2x + 1) = 2x⁴ + x³
x²(2x² - 1) = 2x⁴ - x²
Subtract: (2x⁴ + x³) - (2x⁴ - x²) = x³ + x²

Denominator: (2x² - 1)(2x + 1) = 4x³ + 2x² - 2x - 1

So expression: (x³ + x²)/(4x³ + 2x² - 2x - 1)

Divide num and den by x³: (1 + 1/x)/(4 + 2/x - 2/x² - 1/x³) → 1/4

So limit = 1/4

---

v)

limₓ→∞ [x² - (x⁴ - 1)/(x² - 2)]
Common denominator:

= [x²(x² - 2) - (x⁴ - 1)] / (x² - 2)
= [x⁴ - 2x² - x⁴ + 1] / (x² - 2) = (-2x² + 1)/(x² - 2)

Divide by x²: (-2 + 1/x²)/(1 - 2/x²) → -2/1 = -2

---

w)

limₓ→∞ [(x - 1)¹⁰⁰ (6x + 1)²⁰⁰] / (3x + 5)³⁰⁰
Look at degrees:

Numerator: (x)¹⁰⁰ * (6x)²⁰⁰ = 6²⁰⁰ x³⁰⁰
Denominator: (3x)³⁰⁰ = 3³⁰⁰ x³⁰
So ratio → 6²⁰⁰ / 3³⁰⁰ = (2*3)²⁰⁰ / 3³⁰⁰ = 2²⁰⁰ * 3²⁰⁰ / 3³⁰⁰ = 2²⁰⁰ / 3¹⁰⁰

Which is (2² / 3)¹⁰⁰ = (4/3)¹⁰⁰? No:

2²⁰⁰ / 3¹⁰⁰ = (2²)¹⁰⁰ / 3¹⁰⁰ = (4/3)¹⁰⁰? Wait no:

Actually: 2²⁰⁰ / 3¹⁰⁰ = (2²⁰) / (3¹⁰⁰) — cannot simplify further unless written as (2²/3)^100? No.

Better to leave as (6²⁰⁰)/(3³⁰⁰) = (6/3)²⁰⁰ * 1/3¹⁰⁰? Not helpful.

Actually: 6²⁰⁰ / 3³⁰⁰ = (2·3)²⁰⁰ / 3³⁰⁰ = 2²⁰⁰ · 3²⁰⁰ / 3³⁰⁰ = 2²⁰⁰ / 3¹⁰⁰

Yes. So limit = 2²⁰⁰ / 3¹⁰⁰

But perhaps they want numerical value? No, it’s fine as is. Or write as (2²/3)^100? 2²⁰⁰ / 3¹⁰⁰ = (2²)¹⁰⁰ / 3¹⁰⁰ = (4/3)¹⁰⁰? Wait no:

2²⁰⁰ = (2²)¹⁰⁰ = 4¹⁰⁰, yes! So 4¹⁰⁰ / 3¹⁰⁰ = (4/3)¹⁰⁰

Oh! I missed that.

So: 2²⁰⁰ / 3¹⁰⁰ = (2²)¹⁰⁰ / 3¹⁰⁰ = 4¹⁰⁰ / 3¹⁰⁰ = (4/3)¹⁰⁰

Is that right? 2²⁰⁰ = (2²)¹⁰⁰ = 4¹⁰⁰ — yes. And 3¹⁰⁰ stays. So yes, (4/3)¹⁰⁰

But let me double-check original:

Numerator: (x-1)^100 ~ x^100, (6x+1)^200 ~ (6x)^200 = 6^200 x^200 → total x^{300} * 6^200

Denominator: (3x+5)^300 ~ (3x)^300 = 3^300 x^300

So ratio: 6^200 / 3^300 = (2·3)^200 / 3^300 = 2^200 · 3^200 / 3^300 = 2^200 / 3^100 = (2^2)^100 / 3^100 = 4^100 / 3^100 = (4/3)^100

Yes! So limit = (4/3)^100

---

x)

limₓ→∞ [∜(x⁵) + ∛(x³) + ⁺√(x⁸)] / (x⁴ + 2)
Simplify radicals:

∜(x) = x^{5/4}
∛(x³) = x^{3/3} = x¹
⁺√(x⁸) — assuming sixth root? The symbol is unclear, but likely sixth root since others are 4th, 3rd. In image it might be \sqrt[6]{x^8} — let me assume that.

In the problem: "⁴√x⁵ + ³√x³ + ⁶√x⁸" over "³√(x⁴ + 2)"

So:

Numerator: x^{5/4} + x^{1} + x^{8/6} = x^{1.25} + x + x^{4/3} ≈ x^{4/3} since 4/3≈1.333 > 1.25 and 1

Denominator: ∛(x⁴) = x^{4/3}

So ratio ≈ x^{4/3} / x^{4/3} = 1? But need coefficients.

Actually, leading term in numerator is x^{4/3} from sixth root term? Let's see exponents:

5/4 = 1.25, 1 = 1, 8/6 = 4/3 ≈ 1.333 — so largest is 4/3.

Coefficient of x^{4/3} in numerator: from ⁺√(x⁸) = x^{8/6} = x^{4/3}, coefficient 1.

Denominator: ∛(x⁴ + 2) ~ x^{4/3}

So limit = 1 / 1 = 1

But is there only one term with x^{4/3}? Yes, and others are smaller order.

So limit = 1

---

y)

limₓ→∞ [x²(2x+1)(3x-2)] / [2x²(5x-8)(x+6)]
Cancel x² top and bottom.

Left with: [(2x+1)(3x-2)] / [2(5x-8)(x+6)]

Expand or look at leading terms:

Num: (2x)(3x) = 6x²
Den: 2*(5x)*(x) = 10x²
So 6x² / 10x² = 6/10 = 3/5

Exactly: divide num and den by x²:

[(2 + 1/x)(3 - 2/x)] / [2(5 - 8/x)(1 + 6/x)] → (2*3)/(2*5*1) = 6/10 = 3/5

---

z)

limₓ→∞ [(2x⁸ + 8x⁶ + 6x⁴)/(4x⁸ - x⁶ + 12x⁴)]⁵
Inside: divide num and den by x⁸:

(2 + 8/x² + 6/x⁴)/(4 - 1/x² + 12/x⁴) → 2/4 = 1/2
Then (1/2)^5 = 1/32

---

Z)

limₓ→∞ [(2x - 3)^20 (3x + 2)^30] / (2x + 1)^50
Leading terms:

Num: (2x)^20 * (3x)^30 = 2^20 * 3^30 * x^{50}
Den: (2x)^50 = 2^50 * x^{50}
Ratio: (2^20 * 3^30) / 2^50 = 3^30 / 2^{30} = (3/2)^30

So limit = (3/2)^30

---

Now compiling 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) 4
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) 1/32
Z) (3/2)^30

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