Worksheet displaying 14 trigonometric limit problems with step-by-step solutions.

Here are the step-by-step solutions for each limit problem shown in the worksheet.

1. $\lim_{x \to 0} \frac{\sin^2 x}{x}$

We can rewrite $\sin^2 x$ as $\sin x \cdot \sin x$.
$$ \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \sin x \right) $$
We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \sin x = 0$.
$$ 1 \cdot 0 = 0 $$

2. $\lim_{x \to 0} \frac{\tan 3x}{x}$

Multiply and divide by 3 to match the argument of the tangent function:
$$ \lim_{x \to 0} 3 \cdot \frac{\tan 3x}{3x} $$
Using the standard limit $\lim_{u \to 0} \frac{\tan u}{u} = 1$:
$$ 3 \cdot 1 = 3 $$

3. $\lim_{x \to 0} \frac{5x + \sin 4x}{\sin 7x}$

Divide the numerator and denominator by $x$:
$$ \lim_{x \to 0} \frac{5 + \frac{\sin 4x}{x}}{\frac{\sin 7x}{x}} $$
Adjust the sine terms to use standard limits ($\lim_{x \to 0} \frac{\sin kx}{x} = k$):
Numerator: $5 + 4(\frac{\sin 4x}{4x}) \to 5 + 4(1) = 9$
Denominator: $7(\frac{\sin 7x}{7x}) \to 7(1) = 7$
Result: $\frac{9}{7}$

4. $\lim_{x \to 0} \frac{x - \cos x}{x}$

Substitute $x = 0$ directly:
Numerator: $0 - \cos(0) = 0 - 1 = -1$
Denominator: $0$
Since we have a non-zero number divided by zero, the limit is undefined (approaches infinity). Specifically, from the right it goes to $-\infty$ and from the left to $+\infty$. In many calculus contexts, this is simply stated as undefined or does not exist. However, looking at the pattern of other problems, let's re-read carefully. Ah, usually these problems result in finite numbers. Let's look closer. Is it possible the question implies one-sided? No. Let's assume standard evaluation.
Wait, let's look at problem 4 again. $\frac{x - \cos x}{x} = 1 - \frac{\cos x}{x}$. As $x \to 0$, $\frac{\cos x}{x}$ blows up. The limit Does Not Exist.

*(Self-Correction/Note: In some specific textbook contexts, if there was a typo for $1-\cos x$, the answer would be 0. But strictly as written, it does not exist. I will provide "Does Not Exist" as the mathematically correct answer.)*

5. $\lim_{x \to 0} \frac{2 \cos x - 2}{x}$

Factor out the 2:
$$ 2 \lim_{x \to 0} \frac{\cos x - 1}{x} $$
This is a standard limit identity: $\lim_{x \to 0} \frac{\cos x - 1}{x} = 0$.
$$ 2 \cdot 0 = 0 $$

6. $\lim_{x \to 0} \frac{2 \cos x - 2}{x^2}$

Factor out the 2:
$$ 2 \lim_{x \to 0} \frac{\cos x - 1}{x^2} $$
Standard limit identity: $\lim_{x \to 0} \frac{\cos x - 1}{x^2} = -\frac{1}{2}$.
$$ 2 \cdot \left(-\frac{1}{2}\right) = -1 $$

7. $\lim_{x \to 0} \frac{\sin x (1 - 3x)}{x}$

Split the fraction:
$$ \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot (1 - 3x) \right) $$
As $x \to 0$, $\frac{\sin x}{x} \to 1$ and $(1 - 3x) \to 1$.
$$ 1 \cdot 1 = 1 $$

8. $\lim_{x \to 0} \frac{\cos x (1 - 3x)}{x}$

Substitute $x=0$:
Numerator: $\cos(0)(1 - 0) = 1 \cdot 1 = 1$
Denominator: $0$
Non-zero divided by zero means the limit Does Not Exist (approaches $\infty$).

9. $\lim_{x \to 0} \frac{6x^2 + 3x \sin x}{x^2}$

Divide every term by $x^2$:
$$ \lim_{x \to 0} \left( 6 + \frac{3 \sin x}{x} \right) $$
We know $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
$$ 6 + 3(1) = 9 $$

10. $\lim_{x \to 0} \frac{\cos x}{x}$

Substitute $x=0$:
Numerator: $\cos(0) = 1$
Denominator: $0$
The limit Does Not Exist (approaches $\infty$).

11. $\lim_{x \to 0} \frac{1 - 3x^2 - 3 \cos x + \cos^2 x}{x^2}$

Let's simplify the numerator using Taylor series or algebraic manipulation.
Recall $\cos x \approx 1 - \frac{x^2}{2}$. So $\cos^2 x \approx (1 - \frac{x^2}{2})^2 \approx 1 - x^2$.
Numerator $\approx 1 - 3x^2 - 3(1 - \frac{x^2}{2}) + (1 - x^2)$
$= 1 - 3x^2 - 3 + \frac{3x^2}{2} + 1 - x^2$
$= (1 - 3 + 1) + (-3 + 1.5 - 1)x^2$
$= -1 - 2.5x^2$
As $x \to 0$, the numerator approaches $-1$ and denominator approaches $0$.
Limit Does Not Exist.

*(Alternative Check: Did I copy the question right? Usually these cancel out. Let's check $x \to 0$ value directly. Num: $1 - 0 - 3(1) + 1 = -1$. Denom: $0$. Yes, DNE.)*

12. $\lim_{x \to 0} \frac{1 + \cos x}{\sin x}$

Substitute $x=0$:
Numerator: $1 + \cos(0) = 1 + 1 = 2$
Denominator: $\sin(0) = 0$
The limit Does Not Exist (approaches $\infty$).

13. $\lim_{x \to 0} \frac{1 + \cos x}{x \sin x}$

Substitute $x=0$:
Numerator: $1 + 1 = 2$
Denominator: $0 \cdot 0 = 0$
The limit Does Not Exist.

14. $\lim_{x \to 0} \frac{1 + \cos x}{x^2 + \sin x}$

Substitute $x=0$:
Numerator: $1 + 1 = 2$
Denominator: $0 + 0 = 0$
The limit Does Not Exist.

15. $\lim_{x \to 0} \frac{\sin^2 x}{(x - \sin x)^2}$

Use Taylor series approximations for small $x$:
$\sin x \approx x - \frac{x^3}{6}$
So, $x - \sin x \approx x - (x - \frac{x^3}{6}) = \frac{x^3}{6}$
Denominator becomes $(\frac{x^3}{6})^2 = \frac{x^6}{36}$
Numerator is $\sin^2 x \approx x^2$
Limit is $\lim_{x \to 0} \frac{x^2}{\frac{x^6}{36}} = \lim_{x \to 0} \frac{36}{x^4}$
As $x \to 0$, this goes to infinity.
The limit Does Not Exist (approaches $+\infty$).

16. $\lim_{x \to 0} \frac{\tan 2x}{x}$

Multiply and divide by 2:
$$ \lim_{x \to 0} 2 \cdot \frac{\tan 2x}{2x} $$
Using standard limit $\lim_{u \to 0} \frac{\tan u}{u} = 1$:
$$ 2 \cdot 1 = 2 $$

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Summary Note: Several problems on this sheet (4, 8, 10, 11, 12, 13, 14, 15) result in division by zero where the numerator is not zero. In standard calculus, these limits Do Not Exist (or approach infinity). If your class uses "$\infty$" as an answer for vertical asymptotes approaching positive infinity from both sides, you may write $\infty$ for #15. For the others, the signs differ from left to right, so "Does Not Exist" is the precise answer.

Final Answer:
1. 0
2. 3
3. 9/7
4. Does Not Exist
5. 0
6. -1
7. 1
8. Does Not Exist
9. 9
10. Does Not Exist
11. Does Not Exist
12. Does Not Exist
13. Does Not Exist
14. Does Not Exist
15. Does Not Exist (approaches $+\infty$)
16. 2