Comprehensive set of calculus limit problems covering trigonometric identities and algebraic simplifications.

It looks like you've uploaded a list of 26 limit problems involving trigonometric, logarithmic, and algebraic functions. Since I can't view images directly, but you've provided the text of the limits, I'll solve each one step by step, explaining key concepts and techniques used.

We'll go through each limit using standard calculus tools:

- Standard limits:
- $\lim_{x \to 0} \frac{\sin x}{x} = 1$
- $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
- $\lim_{x \to 0} \frac{\tan x}{x} = 1$
- $\lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}$ for constants $a,b$
- $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$, but $\frac{1 - \cos x}{x^2} \to \frac{1}{2}$
- $\lim_{x \to 0} x \sin\left(\frac{\pi}{x}\right) = 0$ (bounded × infinitesimal)
- $\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$

Let’s now evaluate each one.

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a) $\lim_{x \to 0} \frac{\sin(10x)}{10x}$

This is a standard form:
$$
\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1
$$
So,
$$
\frac{\sin(10x)}{10x} \to 1 \quad \text{as } x \to 0
$$
✔ Answer: 1

---

b) $\lim_{x \to 0} \frac{\sin(3x)}{2x}$

Use: $\frac{\sin(3x)}{3x} \to 1$, so:
$$
\frac{\sin(3x)}{2x} = \frac{3}{2} \cdot \frac{\sin(3x)}{3x} \to \frac{3}{2} \cdot 1 = \frac{3}{2}
$$
✔ Answer: $\frac{3}{2}$

---

c) $\lim_{x \to 0} \frac{\tan(8x)}{x}$

$\tan(8x) \sim 8x$ as $x \to 0$, so:
$$
\frac{\tan(8x)}{x} \sim \frac{8x}{x} = 8
$$
✔ Answer: 8

---

d) $\lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)}$

Use $\sin(kx) \sim kx$:
$$
\frac{\sin(3x)}{\sin(5x)} \sim \frac{3x}{5x} = \frac{3}{5}
$$
✔ Answer: $\frac{3}{5}$

---

e) $\lim_{x \to 0} \frac{\tan(5x)}{\sin(4x)}$

$\tan(5x) \sim 5x$, $\sin(4x) \sim 4x$:
$$
\frac{5x}{4x} = \frac{5}{4}
$$
✔ Answer: $\frac{5}{4}$

---

f) $\lim_{x \to 0} \frac{\tan(5x)}{\tan(6x)}$

$\tan(5x) \sim 5x$, $\tan(6x) \sim 6x$:
$$
\frac{5x}{6x} = \frac{5}{6}
$$
✔ Answer: $\frac{5}{6}$

---

g) $\lim_{x \to 0} \frac{\tan x}{3x}$

$\tan x \sim x$, so:
$$
\frac{x}{3x} = \frac{1}{3}
$$
✔ Answer: $\frac{1}{3}$

---

h) $\lim_{x \to 0} \frac{1 - \cos x}{x}$

We know: $1 - \cos x \sim \frac{x^2}{2}$, so:
$$
\frac{1 - \cos x}{x} \sim \frac{x^2/2}{x} = \frac{x}{2} \to 0
$$
✔ Answer: 0

---

i) $\lim_{x \to 0} \frac{x}{\sqrt{1 - \cos x}}$

Again, $1 - \cos x \sim \frac{x^2}{2}$, so:
$$
\sqrt{1 - \cos x} \sim \sqrt{\frac{x^2}{2}} = \frac{|x|}{\sqrt{2}}
$$
As $x \to 0$, we assume $x > 0$ (limit from right), so $|x| = x$. Then:
$$
\frac{x}{\sqrt{1 - \cos x}} \sim \frac{x}{x/\sqrt{2}} = \sqrt{2}
$$
✔ Answer: $\sqrt{2}$

---

j) $\lim_{x \to 0} \frac{\sin x}{x^3}$

$\sin x \sim x$, so:
$$
\frac{x}{x^3} = \frac{1}{x^2} \to \infty
$$
But since numerator → 0, denominator → 0, but faster? Wait:

Wait: $\sin x \sim x$, so:
$$
\frac{\sin x}{x^3} \sim \frac{x}{x^3} = \frac{1}{x^2} \to \infty
$$
Actually, this diverges to $+\infty$ as $x \to 0^+$, $-\infty$ as $x \to 0^-$, but since it's $1/x^2$, always positive.

But wait: $\sin x$ is odd, $x^3$ is odd ⇒ ratio is even?

No: $\frac{\sin x}{x^3}$: both numerator and denominator are odd ⇒ ratio is even function.

But as $x \to 0$, $\sin x \approx x - \frac{x^3}{6}$, so:
$$
\frac{\sin x}{x^3} \approx \frac{x - x^3/6}{x^3} = \frac{1}{x^2} - \frac{1}{6}
\to \infty
$$
So the limit is infinity.

But let's check: as $x \to 0$, numerator ~ $x$, denominator ~ $x^3$, so overall ~ $1/x^2$ → ∞

✔ Answer: $\infty$ (does not exist in finite sense)

---

k) $\lim_{x \to 0} \frac{1 - \cos(2x)}{x \sin x}$

Use:
- $1 - \cos(2x) = 2\sin^2 x \sim 2x^2$
- $\sin x \sim x$, so $x \sin x \sim x^2$

Thus:
$$
\frac{2x^2}{x^2} = 2
$$
✔ Answer: 2

---

l) $\lim_{x \to 0} \frac{\sin^3(x/2)}{x^3}$

$\sin(x/2) \sim x/2$, so:
$$
\sin^3(x/2) \sim \left(\frac{x}{2}\right)^3 = \frac{x^3}{8}
$$
Then:
$$
\frac{x^3/8}{x^3} = \frac{1}{8}
$$
✔ Answer: $\frac{1}{8}$

---

m) $\lim_{x \to 0} \frac{\sin(4x) + \sin(7x)}{\sin(3x)}$

Use $\sin(kx) \sim kx$:
Numerator: $4x + 7x = 11x$, denominator: $3x$
$$
\frac{11x}{3x} = \frac{11}{3}
$$
✔ Answer: $\frac{11}{3}$

---

n) $\lim_{x \to 0} \frac{\sqrt{1 - \cos x^2}}{1 - \cos x}$

First, $1 - \cos x^2 \sim \frac{(x^2)^2}{2} = \frac{x^4}{2}$, so:
$$
\sqrt{1 - \cos x^2} \sim \sqrt{\frac{x^4}{2}} = \frac{x^2}{\sqrt{2}}
$$
Denominator: $1 - \cos x \sim \frac{x^2}{2}$

So:
$$
\frac{x^2 / \sqrt{2}}{x^2 / 2} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}
$$
✔ Answer: $\sqrt{2}$

---

o) $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}$

Use Taylor series:
- $\tan x = x + \frac{x^3}{3} + O(x^5)$
- $\sin x = x - \frac{x^3}{6} + O(x^5)$

So:
$$
\tan x - \sin x = \left(x + \frac{x^3}{3}\right) - \left(x - \frac{x^3}{6}\right) = \frac{x^3}{3} + \frac{x^3}{6} = \frac{x^3}{2}
$$

Then:
$$
\frac{\tan x - \sin x}{x^3} \to \frac{x^3/2}{x^3} = \frac{1}{2}
$$
✔ Answer: $\frac{1}{2}$

---

p) $\lim_{x \to \infty} x \sin\left(\frac{\pi}{x}\right)$

Let $t = \frac{1}{x}$, then as $x \to \infty$, $t \to 0^+$

So:
$$
x \sin\left(\frac{\pi}{x}\right) = \frac{1}{t} \cdot \sin(\pi t) = \frac{\sin(\pi t)}{t}
\to \pi \cdot \frac{\sin(\pi t)}{\pi t} \to \pi \cdot 1 = \pi
$$
✔ Answer: $\pi$

---

q) $\lim_{x \to -1} \frac{x^3 + 1}{\sin(x+1)}$

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

So:
$$
\frac{(x+1)(x^2 - x + 1)}{\sin(x+1)}
= (x^2 - x + 1) \cdot \frac{x+1}{\sin(x+1)}
$$

As $x \to -1$, $x+1 \to 0$, so $\frac{x+1}{\sin(x+1)} \to 1$

And $x^2 - x + 1$ at $x = -1$: $1 + 1 + 1 = 3$

So limit = $3 \cdot 1 = 3$

✔ Answer: 3

---

r) $\lim_{x \to 0} \frac{|\sin x|}{x}$

We must consider left and right limits.

As $x \to 0^+$: $\sin x > 0$, so $|\sin x| = \sin x$, so:
$$
\frac{\sin x}{x} \to 1
$$

As $x \to 0^-$: $\sin x < 0$, so $|\sin x| = -\sin x$, and $x < 0$, so:
$$
\frac{-\sin x}{x} = \frac{-(-|x|)}{x} = \frac{|x|}{x} = -1 \quad \text{wait! Let's compute:}
$$

Let $x \to 0^-$, say $x = -h$, $h \to 0^+$

Then:
$$
\frac{|\sin(-h)|}{-h} = \frac{|-\sin h|}{-h} = \frac{\sin h}{-h} = -\frac{\sin h}{h} \to -1
$$

So left limit: $-1$, right limit: $1$

Hence, limit does not exist.

But if the problem expects an answer, maybe they want the two-sided limit — which doesn’t exist.

But sometimes people write $|\sin x|/x$ and expect consideration.

✔ Answer: Does not exist (left ≠ right)

---

s) $\lim_{x \to 1} \frac{\tan(x-1)}{\sqrt{x} - 1}$

Let $t = x - 1$, so as $x \to 1$, $t \to 0$

Then:
$$
\frac{\tan t}{\sqrt{1 + t} - 1}
$$

Now rationalize denominator:
$$
\frac{\tan t}{\sqrt{1+t} - 1} \cdot \frac{\sqrt{1+t} + 1}{\sqrt{1+t} + 1} = \frac{\tan t (\sqrt{1+t} + 1)}{(1+t) - 1} = \frac{\tan t (\sqrt{1+t} + 1)}{t}
$$

Now:
$$
\frac{\tan t}{t} \to 1, \quad \sqrt{1+t} + 1 \to 2
$$

So total → $1 \cdot 2 = 2$

✔ Answer: 2

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t) $\lim_{x \to 0} \frac{\sqrt{\cos x} - 1}{\sin^2 x}$

Use approximations:
- $\cos x \sim 1 - \frac{x^2}{2}$, so $\sqrt{\cos x} \sim \sqrt{1 - x^2/2} \sim 1 - \frac{x^2}{4}$ (using $\sqrt{1 - u} \sim 1 - u/2$)
- So $\sqrt{\cos x} - 1 \sim -\frac{x^2}{4}$
- $\sin^2 x \sim x^2$

So:
$$
\frac{-x^2/4}{x^2} = -\frac{1}{4}
$$

✔ Answer: $-\frac{1}{4}$

---

u) $\lim_{x \to 0} \frac{\sin(3x) + \sin(5x)}{\sin(2x)}$

Use sum-to-product or approximation:

$\sin(3x) \sim 3x$, $\sin(5x) \sim 5x$, $\sin(2x) \sim 2x$

So:
$$
\frac{3x + 5x}{2x} = \frac{8x}{2x} = 4
$$

✔ Answer: 4

---

v) $\lim_{x \to 0} \frac{\cos x - \cos^3 x}{x^2}$

Factor numerator:
$$
\cos x - \cos^3 x = \cos x (1 - \cos^2 x) = \cos x \cdot \sin^2 x
$$

So:
$$
\frac{\cos x \cdot \sin^2 x}{x^2} = \cos x \cdot \left(\frac{\sin x}{x}\right)^2 \to 1 \cdot 1^2 = 1
$$

✔ Answer: 1

---

w) $\lim_{x \to 0} x \cot(2x)$

$\cot(2x) = \frac{\cos(2x)}{\sin(2x)}$, so:
$$
x \cdot \frac{\cos(2x)}{\sin(2x)} = \frac{x}{\sin(2x)} \cdot \cos(2x)
$$

Now:
- $\frac{x}{\sin(2x)} = \frac{1}{2} \cdot \frac{2x}{\sin(2x)} \to \frac{1}{2}$
- $\cos(2x) \to 1$

So product → $\frac{1}{2} \cdot 1 = \frac{1}{2}$

✔ Answer: $\frac{1}{2}$

---

x) $\lim_{x \to 0} \sqrt{\frac{1 - \cos x}{x^2}}$

We know:
$$
\frac{1 - \cos x}{x^2} \to \frac{1}{2}
$$

So:
$$
\sqrt{\frac{1 - \cos x}{x^2}} \to \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
$$

✔ Answer: $\frac{\sqrt{2}}{2}$

---

y) $\lim_{x \to 0} \frac{\sqrt{1 - \tan x} - \sqrt{1 + \tan x}}{\sin(2x)}$

Let’s rationalize numerator:

Multiply numerator and denominator by conjugate:
$$
\frac{(\sqrt{1 - \tan x} - \sqrt{1 + \tan x})(\sqrt{1 - \tan x} + \sqrt{1 + \tan x})}{\sin(2x) (\sqrt{1 - \tan x} + \sqrt{1 + \tan x})}
= \frac{(1 - \tan x) - (1 + \tan x)}{\sin(2x) (\cdots)} = \frac{-2 \tan x}{\sin(2x) (\sqrt{1 - \tan x} + \sqrt{1 + \tan x})}
$$

Now:
- $\tan x \sim x$
- $\sin(2x) \sim 2x$
- Denominator: $\sqrt{1 - \tan x} + \sqrt{1 + \tan x} \to \sqrt{1} + \sqrt{1} = 2$

So:
$$
\frac{-2x}{2x \cdot 2} = \frac{-2x}{4x} = -\frac{1}{2}
$$

✔ Answer: $-\frac{1}{2}$

---

z) $\lim_{x \to 0} \frac{\ln x}{\ln(\sin x)}$

As $x \to 0^+$:
- $\ln x \to -\infty$
- $\sin x \sim x$, so $\ln(\sin x) \sim \ln x$

So:
$$
\frac{\ln x}{\ln(\sin x)} \sim \frac{\ln x}{\ln x} = 1
$$

More rigorously:
$$
\frac{\ln x}{\ln(\sin x)} = \frac{\ln x}{\ln(x \cdot \frac{\sin x}{x})} = \frac{\ln x}{\ln x + \ln(\frac{\sin x}{x})}
$$

Now $\frac{\sin x}{x} \to 1$, so $\ln(\frac{\sin x}{x}) \to 0$

So:
$$
\frac{\ln x}{\ln x + o(1)} = \frac{1}{1 + \frac{o(1)}{\ln x}} \to 1
$$

✔ Answer: 1

---

Z) $\lim_{x \to 0} \frac{\ln(\sin 2x)}{\ln(\sin x)}$

As $x \to 0^+$:
- $\sin 2x \sim 2x$, so $\ln(\sin 2x) \sim \ln(2x) = \ln 2 + \ln x$
- $\sin x \sim x$, so $\ln(\sin x) \sim \ln x$

So:
$$
\frac{\ln 2 + \ln x}{\ln x} = \frac{\ln x}{\ln x} + \frac{\ln 2}{\ln x} = 1 + \frac{\ln 2}{\ln x}
$$

As $x \to 0^+$, $\ln x \to -\infty$, so $\frac{\ln 2}{\ln x} \to 0$

Thus, limit → $1 + 0 = 1$

✔ Answer: 1

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## ✔ Final Answers Summary:

| Letter | Answer |
|-------|--------|
| a) | 1 |
| b) | $\frac{3}{2}$ |
| c) | 8 |
| d) | $\frac{3}{5}$ |
| e) | $\frac{5}{4}$ |
| f) | $\frac{5}{6}$ |
| g) | $\frac{1}{3}$ |
| h) | 0 |
| i) | $\sqrt{2}$ |
| j) | $\infty$ |
| k) | 2 |
| l) | $\frac{1}{8}$ |
| m) | $\frac{11}{3}$ |
| n) | $\sqrt{2}$ |
| o) | $\frac{1}{2}$ |
| p) | $\pi$ |
| q) | 3 |
| r) | Does not exist |
| s) | 2 |
| t) | $-\frac{1}{4}$ |
| u) | 4 |
| v) | 1 |
| w) | $\frac{1}{2}$ |
| x) | $\frac{\sqrt{2}}{2}$ |
| y) | $-\frac{1}{2}$ |
| z) | 1 |
| Z) | 1 |

Let me know if you'd like detailed steps for any specific one!