Comprehensive set of limit problems in calculus, arranged in a grid format, covering a range of functions including polynomials, exponentials, logarithms, and trigonometric expressions.

Let’s solve each limit one by one. We’ll use basic limit rules, factoring, L’Hôpital’s Rule when needed (0/0 or ∞/∞ forms), and known standard limits like:

- limₓ→₀ sinx/x = 1
- limₓ→₀ (eˣ - 1)/x = 1
- limₓ→₀ ln(1+x)/x = 1
- limₓ→₀ arctanx/x = 1

We’ll go in order from a) to Z).

---

a) limₓ→₁ (x³ - 2x² + x)/(x² - 1)

Factor numerator and denominator:

Numerator: x(x² - 2x + 1) = x(x - 1)²
Denominator: (x - 1)(x + 1)

So expression becomes: [x(x - 1)²] / [(x - 1)(x + 1)] = [x(x - 1)] / (x + 1) for x ≠ 1

Now plug in x = 1: [1*(0)] / 2 = 0

✔ Answer: 0

---

b) limₓ→₃ (x³ - 9x)/(x⁴ - 3x³ - x + 3)

Factor numerator: x(x² - 9) = x(x - 3)(x + 3)

Denominator: group terms: (x⁴ - 3x³) + (-x + 3) = x³(x - 3) -1(x - 3) = (x³ - 1)(x - 3)

Wait — actually: x⁴ - 3x³ - x + 3 = x³(x - 3) -1(x - 3) = (x³ - 1)(x - 3)

But x³ - 1 = (x - 1)(x² + x + 1)

So overall: denominator = (x - 1)(x² + x + 1)(x - 3)

Numerator: x(x - 3)(x + 3)

Cancel (x - 3):

Left with: [x(x + 3)] / [(x - 1)(x² + x + 1)]

Plug in x = 3: [3*6] / [2 * (9 + 3 + 1)] = 18 / (2 * 13) = 18/26 = 9/13

✔ Answer: 9/13

---

c) limₓ→₁ (∛(8x) - 2x) / (∜x - x)

First, simplify radicals:

∛(8x) = 2 ∛x
∜x = x^{1/4}

So: [2 x^{1/3} - 2x] / [x^{1/4} - x] = 2[x^{1/3} - x] / [x^{1/4} - x]

Factor out x from numerator and denominator? Not directly. Let’s set t = x^{1/12}, so that:

x = t¹²
x^{1/3} = t⁴
x^{1/4} = t³

Then as x → 1, t → 1

Expression becomes:

2[t⁴ - t¹²] / [t³ - t¹²] = 2t⁴(1 - t⁸) / [t³(1 - t⁹)] = 2t (1 - t⁸)/(1 - t⁹)

As t → 1, this is 0/0 → use L’Hôpital or factor.

Note: 1 - t⁸ = (1 - t)(1 + t + ... + t⁷)
1 - t⁹ = (1 - t)(1 + t + ... + t⁸)

So cancel (1 - t):

= 2t * [sum_{k=0}^7 t^k] / [sum_{k=0}^8 t^k]

At t = 1: 2*1 * (8) / (9) = 16/9

✔ Answer: 16/9

---

d) limₓ→₁ (√(x+1) - √2) / (x² - 1)

Multiply numerator and denominator by conjugate of numerator: √(x+1) + √2

Numerator becomes: (x+1) - 2 = x - 1

Denominator: (x² - 1)(√(x+1) + √2) = (x - 1)(x + 1)(√(x+1) + √2)

Cancel (x - 1):

Left with: 1 / [(x + 1)(√(x+1) + √2)]

Plug in x = 1: 1 / [2 * (√2 + √2)] = 1 / [2 * 2√2] = 1/(4√2) = √2/8 after rationalizing

✔ Answer: √2 / 8

---

e) limₓ→₀ (5ˣ - 1)/x

Standard limit: limₓ→₀ (aˣ - 1)/x = ln a

So here: ln 5

✔ Answer: ln 5

---

f) limₓ→₁ (3x³ - 3)/(3ˣ - 3)

Factor numerator: 3(x³ - 1) = 3(x - 1)(x² + x + 1)

Denominator: 3ˣ - 3 = 3(3^{x-1} - 1)

Set u = x - 1 → as x→1, u→0

Then: 3(u+1)³ - 3 = 3[(u+1)³ - 1] = 3[u³ + 3u² + 3u] = 3u(u² + 3u + 3)

Denominator: 3^{u+1} - 3 = 3·3ᵘ - 3 = 3(3ᵘ - 1)

So expression: [3u(u² + 3u + 3)] / [3(3ᵘ - 1)] = u(u² + 3u + 3)/(3ᵘ - 1)

As u→0: numerator ~ u·3, denominator ~ u ln3 (since 3ᵘ - 1 ~ u ln3)

So limit = 3 / ln3

Wait — let's do it properly:

limᵤ→₀ u / (3ᵘ - 1) * (u² + 3u + 3) = [1 / ln3] * 3 = 3 / ln3

Because limᵤ→₀ (3ᵘ - 1)/u = ln3 → so lim u/(3ᵘ - 1) = 1/ln3

And (u² + 3u + 3) → 3

So total: 3 / ln3

✔ Answer: 3 / ln 3

---

g) limₓ→₀ (eˣ - e⁻)/x

This is 2 sinh(x)/x → but we can write as:

(eˣ - 1)/x - (e⁻ˣ - 1)/x → as x→0, first term → 1, second term: let u=-x → (eᵘ - 1)/(-u) → -1, so minus that is +1 → total 1 - (-1)? Wait no:

Actually: (eˣ - e⁻ˣ)/x = [ (eˣ - 1) - (e⁻ˣ - 1) ] / x = (eˣ - 1)/x - (e⁻ˣ - 1)/x

As x→0: first part → 1, second part: let u = -x → (eᵘ - 1)/(-u) → -1, so -( -1 ) = +1? No:

Wait: (e⁻ˣ - 1)/x = [e⁻ˣ - 1]/x → as x→0, this is same as lim_{u→0} (eᵘ - 1)/(-u) = - lim (eᵘ - 1)/u = -1

So overall: 1 - (-1) = 2? No:

Original: (eˣ - e⁻ˣ)/x = A - B, where A = (eˣ - 1)/x → 1, B = (e⁻ˣ - 1)/x → -1

So A - B = 1 - (-1) = 2

Alternatively, derivative of eˣ - e⁻ˣ at 0 is e⁰ + e⁰ = 2, and since it's 0/0, L’Hôpital gives 2.

✔ Answer: 2

---

h) limₓ→₀ (eˣ - 1)/x²

As x→0, numerator ~ x, denominator x² → so ~ 1/x → goes to ∞

More precisely: (eˣ - 1)/x² = [x + x²/2 + ...]/x² = 1/x + 1/2 + ... → as x→0+, → +∞; x→0-, → -∞ → so limit does not exist? But usually in such contexts, if not specified, we say DNE or ±∞.

But looking at the form: it's 0/0, apply L’Hôpital:

Derivative num: eˣ, den: 2x → still 1/0 → ∞

So limit is ∞ (but sign depends on side). Since problem doesn't specify, and often in such lists they expect "does not exist" or "infinity". But let's check behavior.

Actually, for x>0 small, positive; x<0 small, negative → so two-sided limit does not exist.

But perhaps in context, they want to say ∞? I think better to say does not exist. However, many textbooks would write ∞ if both sides go to same infinity, but here left is -∞, right is +∞.

Wait: (eˣ - 1) for x<0 is negative, x² always positive → so for x<0, expression is negative → so left limit -∞, right limit +∞ → so two-sided limit does not exist.

But let me double-check with L’Hôpital once:

After first L’Hôpital: eˣ / (2x) → as x→0+, → +∞; x→0- → -∞ → so yes, DNE.

However, sometimes problems assume x→0+ unless specified. But here it's written x→0, so probably DNE.

But looking at other problems, maybe they expect ∞? I think safest is to say the limit does not exist.

But let's see answer format — perhaps they want a number. Maybe I made mistake.

Another way: Taylor series: eˣ = 1 + x + x²/2 + O(x³) → eˣ - 1 = x + x²/2 + O(x³) → divided by x²: 1/x + 1/2 + O(x) → as x→0, blows up, and sign changes.

So I think correct answer is: does not exist

But since all others are numbers, perhaps I should write ∞? No, because left and right differ.

Perhaps the problem intends x→0+, but it's not specified.

Looking back at original image, it's written as x→0, so I'll go with does not exist. But to match format, maybe write "DNE".

But let's proceed and come back.

Actually, in many calculus courses, for such limits, if it goes to ±∞ depending on side, they say the limit does not exist. So I'll put DNE.

But let's check online or standard — actually, for (eˣ - 1)/x², it's well-known that lim_{x→0} does not exist.

✔ Answer: Does not exist

But to be consistent with numerical answers, perhaps I missed something. Another thought: maybe they want the value using L’Hôpital twice? First derivative: eˣ / (2x) → still indeterminate? No, eˣ→1, 2x→0, so not indeterminate anymore — it's 1/0, which is infinite.

So I think DNE is correct.

But let's move on and return.

---

i) limₓ→π/3 (1 - 2 cos x)/(π - 3x)

Let u = x - π/3 → as x→π/3, u→0

Then x = u + π/3

cos x = cos(u + π/3) = cos u cos π/3 - sin u sin π/3 = (1/2)cos u - (√3/2) sin u

So 1 - 2 cos x = 1 - 2[ (1/2)cos u - (√3/2) sin u ] = 1 - cos u + √3 sin u

Denominator: π - 3x = π - 3(u + π/3) = π - 3u - π = -3u

So expression: [1 - cos u + √3 sin u] / (-3u)

As u→0: 1 - cos u ~ u²/2, sin u ~ u

So numerator ~ u²/2 + √3 u

Divided by -3u: ~ (√3 u) / (-3u) + (u²/2)/(-3u) = -√3/3 - u/6 → as u→0, → -√3/3

More rigorously:

= [1 - cos u]/(-3u) + [√3 sin u]/(-3u) = - (1 - cos u)/(3u) - (√3/3) (sin u)/u

As u→0: (1 - cos u)/u → 0, (sin u)/u → 1

So limit = 0 - (√3/3)*1 = -√3/3

✔ Answer: -√3 / 3

---

j) limₓ→₀ sinx / (eˣ - 1)

Both →0, so L’Hôpital: cosx / eˣ → 1/1 = 1

Or use standard limits: sinx/x →1, (eˣ -1)/x →1, so ratio →1

✔ Answer: 1

---

k) limₓ→₀ arctan(3x) / arcsin(2x)

As x→0, arctan(3x) ~ 3x, arcsin(2x) ~ 2x, so ratio ~ 3x/2x = 3/2

More formally: divide numerator and denominator by x:

[arctan(3x)/x] / [arcsin(2x)/x] = [3 * arctan(3x)/(3x)] / [2 * arcsin(2x)/(2x)] → 3*1 / (2*1) = 3/2

✔ Answer: 3/2

---

l) limₓ→₀ (eˣ - e⁻ˣ)/(2x - sin x)

Numerator: eˣ - e⁻ ~ 2x (from earlier, or Taylor: 2x + 2x³/6 + ...)

Denominator: 2x - sin x ~ 2x - (x - x³/6) = x + x³/6

So ratio ~ (2x) / x = 2 as x→0

More precisely: use L’Hôpital since 0/0

Num deriv: eˣ + e⁻ˣ
Den deriv: 2 - cos x

At x=0: (1+1)/(2-1) = 2/1 = 2

✔ Answer: 2

---

m) limₓ→₀ (x³ + πx) / sin(3x)

Factor x: x(x² + π) / sin(3x)

= [x / sin(3x)] * (x² + π) = [1/3] * [3x / sin(3x)] * (x² + π)

As x→0, 3x/sin(3x) →1, x²+π→π, so limit = (1/3)*1*π = π/3

✔ Answer: π/3

---

n) limₓ→₀ ln(1+4x)/(3ˣ - 1)

As x→0, ln(1+4x) ~ 4x, 3ˣ - 1 ~ x ln3

So ratio ~ 4x / (x ln3) = 4 / ln3

L’Hôpital: num deriv: 4/(1+4x) →4, den deriv: 3ˣ ln3 → ln3, so 4/ln3

✔ Answer: 4 / ln 3

---

o) limₓ→₀ sin(4x) / ln(1 + sin x)

As x→0, sin(4x) ~ 4x, ln(1 + sin x) ~ sin x ~ x, so ratio ~ 4x/x = 4

L’Hôpital: num deriv: 4 cos(4x) →4, den deriv: [1/(1+sin x)] * cos x → 1/1 * 1 =1, so 4/1=4

✔ Answer: 4

---

p) limₓ→₀ 3 ln(1 - 2x) / (2 arctan(3x))

As x→0, ln(1-2x) ~ -2x, arctan(3x) ~ 3x

So numerator: 3*(-2x) = -6x, denominator: 2*3x = 6x, ratio = -6x/6x = -1

L’Hôpital: num deriv: 3 * [1/(1-2x)] * (-2) = -6/(1-2x) → -6
Den deriv: 2 * [1/(1+(3x)^2)] * 3 = 6/(1+9x²) →6
So -6/6 = -1

✔ Answer: -1

---

q) limₓ→₀ (arctan x + x²) / (2^{3x} - 3^{2x})

Note: 2^{3x} = (2^3)^x = 8^x, 3^{2x} = 9^x

So denominator: 8^x - 9^x

As x→0, numerator: arctan x + x² ~ x + x² →0
Denominator: 8^x - 9^x →1-1=0 → 0/0

Use L’Hôpital:

Num deriv: 1/(1+x²) + 2x →1 at x=0
Den deriv: 8^x ln8 - 9^x ln9 → ln8 - ln9 = ln(8/9)

So limit = 1 / ln(8/9) = 1 / (ln8 - ln9) = 1 / (3ln2 - 2ln3)

But perhaps leave as 1 / ln(8/9) or simplify.

Since 8/9 <1, ln(8/9)<0, so negative.

But let's compute numerically later if needed.

We can write as -1 / ln(9/8)

But I think 1 / ln(8/9) is fine, but perhaps simplify.

Note: ln(8/9) = ln8 - ln9 = 3ln2 - 2ln3

But maybe leave as is.

Another way: factor out 9^x or something, but not necessary.

So answer is 1 / (ln8 - ln9) = 1 / ln(8/9)

But to make it nicer, note that ln(8/9) = - ln(9/8), so = -1 / ln(9/8)

I think either is acceptable, but perhaps write as \frac{1}{\ln \frac{8}{9}} or \frac{-1}{\ln \frac{9}{8}}

In boxed answer, perhaps simplify.

Let me calculate numerical value to verify: ln(8/9) ≈ ln(0.888...) ≈ -0.1178, so 1/-0.1178 ≈ -8.49, and -1/ln(9/8) = -1/ln(1.125) ≈ -1/0.1178 ≈ -8.49, same.

But for exact answer, I'll write \frac{1}{\ln \frac{8}{9}} or better \frac{-1}{\ln \frac{9}{8}}

Since 9/8 >1, ln positive, so negative sign makes sense.

I think \frac{-1}{\ln \frac{9}{8}} is cleaner.

But let's see if we can write in terms of logs.

Note that 8=2^3, 9=3^2, so ln(9/8) = 2ln3 - 3ln2

So answer: -1 / (2ln3 - 3ln2)

But perhaps leave as is.

I think for now, I'll box \frac{1}{\ln \frac{8}{9}} but it's negative, so better to write explicitly.

In the calculation, we have limit = 1 / (ln8 - ln9) = 1 / ln(8/9)

And since 8/9 <1, it's negative, which is fine.

But to avoid confusion, I'll write it as \frac{1}{\ln \frac{8}{9}}

Or perhaps the problem expects simplified form.

Another approach: write denominator as e^{x ln8} - e^{x ln9} ~ x(ln8 - ln9) as x→0, since e^{ax} -1 ~ ax, but here it's e^{a x} - e^{b x} = e^{b x} (e^{(a-b)x} -1) ~ 1 * (a-b)x if b=0, but here both non-zero.

General: e^{A x} - e^{B x} ~ (A - B)x as x→0? Let's see: derivative at 0 is A e^{0} - B e^{0} = A-B, and value at 0 is 0, so yes, by definition of derivative, [f(x)-f(0)]/x → f'(0), so e^{A x} - e^{B x} ~ (A - B)x

Here A=ln8, B=ln9, so ~ (ln8 - ln9)x

Numerator ~ x (since arctan x ~ x, x² negligible)

So ratio ~ x / [(ln8 - ln9)x] = 1/(ln8 - ln9) = 1/ln(8/9)

Same as before.

So I'll go with that.

✔ Answer: \frac{1}{\ln \frac{8}{9}} or equivalently -\frac{1}{\ln \frac{9}{8}}

I think the latter is better because ln(9/8)>0.

So \boxed{ -\frac{1}{\ln \frac{9}{8}} }

But let's confirm with numbers: ln(9/8) = ln1.125 ≈ 0.117783, so -1/0.117783 ≈ -8.49, and from direct substitution small x, say x=0.001, num= arctan(0.001)+0.000001≈0.001, den=8^{0.001}-9^{0.001} = e^{0.001*ln8} - e^{0.001*ln9} ≈ (1 + 0.001*ln8) - (1 + 0.001*ln9) = 0.001(ln8 - ln9) = 0.001*ln(8/9)≈0.001*(-0.117783) = -0.000117783, so ratio 0.001 / (-0.000117783) ≈ -8.49, yes.

So answer is -1 / ln(9/8)

✔ Final for q: -\frac{1}{\ln \frac{9}{8}}

---

r) limₓ→π/2 arctan(x - π/2) / (π - 2x)

Let u = x - π/2 → as x→π/2, u→0

Then x = u + π/2

Numerator: arctan(u)

Denominator: π - 2(u + π/2) = π - 2u - π = -2u

So expression: arctan(u) / (-2u) = -1/2 * [arctan(u)/u] → -1/2 * 1 = -1/2

✔ Answer: -1/2

---

s) limₓ→₀ (e^{3x} - e^{-2x}) / (2 arcsin x - sin x)

As x→0, num: e^{3x} - e^{-2x} ~ (1+3x) - (1-2x) = 5x (first order)

Den: 2 arcsin x - sin x ~ 2x - x = x (since arcsin x ~ x, sin x ~ x)

So ratio ~ 5x / x = 5

L’Hôpital: num deriv: 3e^{3x} + 2e^{-2x} → 3+2=5
Den deriv: 2 / sqrt(1-x²) - cos x → 2/1 - 1 =1
So 5/1=5

✔ Answer: 5

---

t) limₓ→₀ ln(cos 3x) / arctan(4x)

As x→0, cos3x ~ 1 - (9x²)/2, so ln(cos3x) ~ ln(1 - 9x²/2) ~ -9x²/2

Arctan(4x) ~ 4x

So ratio ~ (-9x²/2) / (4x) = -9x/8 → 0 as x→0

More rigorously: L’Hôpital might be messy, but since num ~ x², den ~ x, so overall ~ x →0

Confirm: num is O(x²), den is O(x), so limit 0.

✔ Answer: 0

---

u) limₓ→₀ [(1+x)² - (1+2x)] / (x² + 4x³)

Simplify numerator: (1 + 2x + x²) - 1 - 2x = x²

Denominator: x²(1 + 4x)

So expression: x² / [x²(1 + 4x)] = 1/(1+4x) → 1 as x→0

✔ Answer: 1

---

v) limₓ→-1 (x² + 3x + 2)² / (x³ - 3x - 2)

First, factor numerator: x² + 3x + 2 = (x+1)(x+2), so squared: (x+1)² (x+2)²

Denominator: x³ - 3x - 2, try roots: possible ±1,±2

x=-1: (-1) -3(-1) -2 = -1 +3 -2=0, so (x+1) is factor

Divide: x³ - 3x -2 ÷ (x+1)

Using polynomial division or synthetic: -1 | 1 0 -3 -2
Bring down 1, multiply -1, add to 0 → -1, multiply -1 →1, add to -3→-2, multiply -1→2, add to -2→0

So quotient: x² - x -2 = (x-2)(x+1)

Thus denominator: (x+1)(x² - x -2) = (x+1)² (x-2)

So overall: [ (x+1)² (x+2)² ] / [ (x+1)² (x-2) ] = (x+2)² / (x-2) for x≠-1

Plug in x=-1: (1)² / (-3) = 1/-3 = -1/3

✔ Answer: -1/3

---

w) limₓ→₀ (1 - cos⁴ x) / (4x²)

Note: 1 - cos⁴ x = (1 - cos² x)(1 + cos² x) = sin² x (1 + cos² x)

So expression: [sin² x (1 + cos² x)] / (4x²) = (1/4) (sin x / x)² (1 + cos² x)

As x→0: (sin x / x)² →1, 1 + cos² x →1+1=2, so (1/4)*1*2 = 1/2

✔ Answer: 1/2

---

x) limₓ→₀ (1 - cos x) / (2x sin x)

As x→0, 1 - cos x ~ x²/2, sin x ~ x, so denominator 2x * x = 2x²

Ratio: (x²/2) / (2x²) = (1/2)/2 = 1/4

L’Hôpital: num deriv: sin x, den deriv: 2[sin x + x cos x] (product rule)

At x=0: sin0=0, den: 2[0 + 0*1]=0 → still 0/0

Second L’Hôpital: num deriv: cos x, den deriv: 2[cos x + cos x - x sin x] = 2[2cos x - x sin x]

At x=0: cos0=1, den: 2[2*1 - 0] =4, so 1/4

✔ Answer: 1/4

---

y) limₓ→₀ (x - sin x) / (eˣ - e⁻ˣ - 2x)

As x→0, num: x - sin x ~ x - (x - x³/6) = x³/6

Den: eˣ - e⁻ˣ - 2x ~ [x + x³/6] - [-x + (-x)³/6 wait no

eˣ = 1 + x + x²/2 + x³/6 + O(x⁴)
e⁻ˣ = 1 - x + x²/2 - x³/6 + O(x⁴)
So eˣ - e⁻ˣ = [1+x+x²/2+x³/6] - [1-x+x²/2-x³/6] + O(x⁴) = 2x + 2x³/6 + O(x⁴) = 2x + x³/3 + O(x⁴)

Then eˣ - e⁻ˣ - 2x = x³/3 + O(x⁴)

Numerator: x - sin x = x - (x - x³/6 + O(x⁵)) = x³/6 + O(x⁵)

So ratio: (x³/6) / (x³/3) = (1/6)/(1/3) = 1/2

L’Hôpital three times would work, but this is faster.

✔ Answer: 1/2

---

z) limₓ→₀ (e^{3x} - 3x - 1) / sin² x

As x→0, num: e^{3x} - 3x -1 ~ [1 + 3x + (9x²)/2 + ...] - 3x -1 = (9x²)/2 + O(x³)

Den: sin² x ~ x²

So ratio ~ (9x²/2) / x² = 9/2

L’Hôpital: num deriv: 3e^{3x} - 3, den deriv: 2 sin x cos x = sin 2x

At x=0: 3-3=0, sin0=0 → still 0/0

Second L’Hôpital: num deriv: 9e^{3x}, den deriv: 2 cos 2x

At x=0: 9*1 / (2*1) = 9/2

✔ Answer: 9/2

---

Z) limₓ→₀ x³ / (x - arctan x)

As x→0, arctan x ~ x - x³/3 + O(x⁵)

So denominator: x - (x - x³/3) = x³/3 + O(x⁵)

Numerator: x³

So ratio: x³ / (x³/3) = 3

L’Hôpital: num deriv: 3x², den deriv: 1 - 1/(1+x²) = x²/(1+x²)

So 3x² / [x²/(1+x²)] = 3(1+x²) → 3 as x→0

✔ Answer: 3

---

Now back to h) limₓ→₀ (eˣ - 1)/x²

As discussed, this is 0/0, L’Hôpital: eˣ / (2x) → as x→0+, ∞; x→0-, -∞ → so two-sided limit does not exist.

But in some contexts, they might expect ∞, but strictly speaking, it does not exist.

However, looking at the pattern, all others have finite answers except possibly this.

Perhaps I miscalculated.

Another thought: maybe they mean x→0+, but it's not specified.

In the original image, it's written as x→0, so I think DNE is correct.

But to match the format, and since other problems have answers, perhaps I should write "does not exist" or "undefined".

But let's see if there's a trick.

Note that (eˣ - 1)/x² = [ (eˣ - 1)/x ] / x → 1/x as x→0, which goes to ±∞.

So I think it's safe to say the limit does not exist.

For the sake of completing, I'll put DNE.

But in the final answer list, I'll write it as such.

Now, compiling all answers:

a) 0
b) 9/13
c) 16/9
d) √2 / 8
e) ln 5
f) 3 / ln 3
g) 2
h) Does not exist
i) -√3 / 3
j) 1
k) 3/2
l) 2
m) π/3
n) 4 / ln 3
o) 4
p) -1
q) -1 / ln(9/8)
r) -1/2
s) 5
t) 0
u) 1
v) -1/3
w) 1/2
x) 1/4
y) 1/2
z) 9/2
Z) 3

For h, I'll write "does not exist", but perhaps in boxed, we can write DNE.

Since the instruction is to provide final answer, and for h it's special, I'll include it.

Now for the response, since ExplainAnswer is false, etc., I just need to provide the final answers in order.

The user said "solve the problem accurately", and the problem is to find all these limits.

So I'll list them.

But in the format, it says "Final Answer: [provide only the final correct answer]", but since there are many, I need to list all.

Probably, I should list them as a) to Z).

So let me write the final answers clearly.

Note: for q, I have -1 / ln(9/8), which is the same as 1 / ln(8/9), but I'll use the form with positive log.

Similarly, for others.

Also, for d, √2 / 8 is fine.

Now, for h, I'll write "undefined" or "does not exist". To be precise, "the limit does not exist".

But in math contests, sometimes they write \infty if it goes to infinity, but here it's signed.

I recall that in some textbooks, for (e^x -1)/x^2, they say the limit is \infty if considering x->0+, but here it's two-sided.

Given that, and to be accurate, I'll state it does not exist.

But let's double-check with a reliable source in mind: yes, standard result is that lim_{x->0} (e^x -1)/x^2 does not exist because left and right limits are -∞ and +∞.

So I'll keep it.

Now, for the final answer section, I'll list all.

Since the user might expect a single box, but there are many parts, I think I need to provide all.

Perhaps the "problem" is to solve all, so final answer is the list.

So I'll write:

Final Answer:
a) 0
b) \frac{9}{13}
c) \frac{16}{9}
d) \frac{\sqrt{2}}{8}
e) \ln 5
f) \frac{3}{\ln 3}
g) 2
h) \text{does not exist}
i) -\frac{\sqrt{3}}{3}
j) 1
k) \frac{3}{2}
l) 2
m) \frac{\pi}{3}
n) \frac{4}{\ln 3}
o) 4
p) -1
q) -\frac{1}{\ln \frac{9}{8}}
r) -\frac{1}{2}
s) 5
t) 0
u) 1
v) -\frac{1}{3}
w) \frac{1}{2}
x) \frac{1}{4}
y) \frac{1}{2}
z) \frac{9}{2}
Z) 3

For q, I can write it as \frac{1}{\ln \frac{8}{9}}, but I think -\frac{1}{\ln \frac{9}{8}} is better.

Some might prefer to write in terms of natural logs without fractions, but it's fine.

Now, I think that's all.

But let me verify one more: for c, I got 16/9, let me quick check.

c) lim x->1 (∛(8x) - 2x) / (∜x - x) = (2 x^{1/3} - 2x) / (x^{1/4} - x)

At x=1, 2-2=0, 1-1=0.

With t=x^{1/12}, x=t^{12}, then 2(t^4 - t^{12}) / (t^3 - t^{12}) = 2t^4(1-t^8)/(t^3(1-t^9)) = 2t (1-t^8)/(1-t^9)

As t->1, (1-t^8)/(1-t^9) = [ (1-t)(1+t+...+t^7) ] / [ (1-t)(1+t+...+t^8) ] = sum_{k=0}^7 t^k / sum_{k=0}^8 t^k -> 8/9

Then 2t * 8/9 -> 2*1*8/9=16/9, yes.

Good.

For h, I think it's correct.

So I'll proceed.