\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

The argument inside sine must match the denominator.

\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \]

\[ \lim_{x \to 0} \frac{\sin(2x)}{x} \]

We need to match the denominator with the argument of the sine function. The argument is \(2x\), but the denominator is just \(x\). To fix this, we'll multiply the numerator and denominator by 2.

\[ \lim_{x \to 0} \frac{2 \cdot \sin(2x)}{2 \cdot x} \]

We can move the constant 2 outside the limit.

\[ 2 \cdot \lim_{x \to 0} \frac{\sin(2x)}{2x} \]

Now the argument and the denominator match. By Rule 1, the limit of this fraction is 1.

\[ 2 \cdot (1) = 2 \]

We want to find \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} \). Since direct substitution gives the indeterminate form \( \frac{0}{0} \), we can multiply the numerator and denominator by the conjugate of the numerator, which is \( 1 + \cos x \).

\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \frac{1 + \cos x}{1 + \cos x} \]

Using the identity \( (a-b)(a+b) = a^2 - b^2 \), the numerator becomes \( 1 - \cos^2 x \).

\[ \lim_{x \to 0} \frac{1 - \cos^2 x}{x^2(1 + \cos x)} \]

By the Pythagorean identity, \( 1 - \cos^2 x = \sin^2 x \), so we have:

\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2(1 + \cos x)} \]

We can rewrite this expression to apply the special trigonometric limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{\sin x}{x} \cdot \frac{1}{1 + \cos x} \right) \]

Applying the limits to each factor, we get:

\[ \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0} \frac{1}{1 + \cos x} \right) = (1) \cdot (1) \cdot \frac{1}{1 + \cos(0)} = 1 \cdot \frac{1}{1+1} = \frac{1}{2} \]

\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]

We need to make the denominator match the argument of the cosine function. The argument is \(2x\), but we have \(x^2\) in the denominator. Let's rewrite the expression to match the form of Rule 2.

To get \((2x)^2\) in the denominator, we need to multiply by \(2^2 = 4\).

\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \cdot \frac{4}{4} \]

Rearrange the terms to get the correct denominator.

\[ \lim_{x \to 0} 4 \cdot \frac{1 - \cos(2x)}{4x^2} = \lim_{x \to 0} 4 \cdot \frac{1 - \cos(2x)}{(2x)^2} \]

Now the argument and the denominator match. By a variation of Rule 2, \( \lim_{u \to 0} \frac{1 - \cos(u)}{u^2} = \frac{1}{2} \). Therefore, the limit is:

\[ 4 \cdot \frac{1}{2} = 2 \]

\[ \lim_{x \to 1} \frac{\sin(\pi(x-1))}{x-1} \]

First, recognize that if we let \( u = x-1 \), then as \( x \to 1 \), we have \( u \to 0 \). This allows us to use Rule 1.

\[ \lim_{u \to 0} \frac{\sin(\pi u)}{u} \]

Now we need to match the argument and the denominator. The argument is \(\pi u\), so we need \(\pi u\) in the denominator.

\[ \lim_{u \to 0} \frac{\pi \cdot \sin(\pi u)}{\pi \cdot u} \]

We can move the constant \(\pi\) outside the limit.

\[ \pi \cdot \lim_{u \to 0} \frac{\sin(\pi u)}{\pi u} \]

By Rule 1, the limit of the fraction is 1.

\[ \pi \cdot (1) = \pi \]

We want to find \( \lim_{x \to 0} \frac{\sin x}{1 - \cos x} \). Since direct substitution results in the indeterminate form \( \frac{0}{0} \), we can multiply the numerator and denominator by the conjugate of the denominator, \( 1 + \cos x \).

\[ \lim_{x \to 0} \frac{\sin x}{1 - \cos x} \cdot \frac{1 + \cos x}{1 + \cos x} = \lim_{x \to 0} \frac{\sin x(1 + \cos x)}{1 - \cos^2 x} \]

Using the identity \( 1 - \cos^2 x = \sin^2 x \), we can simplify the expression:

\[ \lim_{x \to 0} \frac{\sin x(1 + \cos x)}{\sin^2 x} = \lim_{x \to 0} \frac{1 + \cos x}{\sin x} \]

Now, as \( x \to 0 \), the numerator \( 1 + \cos x \) approaches \( 1 + \cos(0) = 2 \). The denominator \( \sin x \) approaches 0. Since the denominator is approaching zero while the numerator is approaching a non-zero number, the limit goes to infinity.

More specifically, we can consider the one-sided limits:

• As \( x \to 0^+ \), \( \sin x \) is positive, so the limit is \( +\infty \).
• As \( x \to 0^- \), \( \sin x \) is negative, so the limit is \( -\infty \).

Since the left-hand and right-hand limits do not agree, the two-sided limit **does not exist**.

\[ \lim_{x \to 0} \frac{\sin(3x)}{\sin(7x)} \]

We can rewrite this expression as:

\[ \lim_{x \to 0} \frac{\sin(3x)}{\sin(7x)} = \lim_{x \to 0} \frac{\sin(3x)/x}{\sin(7x)/x} \]

Now we can apply Rule 1 to both numerator and denominator:

\[ = \frac{\lim_{x \to 0} \frac{\sin(3x)}{x}}{\lim_{x \to 0} \frac{\sin(7x)}{x}} = \frac{3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x}}{7 \cdot \lim_{x \to 0} \frac{\sin(7x)}{7x}} = \frac{3 \cdot 1}{7 \cdot 1} = \frac{3}{7} \]

\[ \lim_{x \to 0} \frac{\tan(4x)}{\sin(6x)} \]

We can rewrite tangent in terms of sine and cosine:

\[ \lim_{x \to 0} \frac{\tan(4x)}{\sin(6x)} = \lim_{x \to 0} \frac{\sin(4x)/\cos(4x)}{\sin(6x)} = \lim_{x \to 0} \frac{\sin(4x)}{\cos(4x) \cdot \sin(6x)} \]

Now we can separate this into:

\[ = \lim_{x \to 0} \frac{\sin(4x)}{\sin(6x)} \cdot \frac{1}{\cos(4x)} \]

From Example F, we know that \( \lim_{x \to 0} \frac{\sin(4x)}{\sin(6x)} = \frac{4}{6} = \frac{2}{3} \). Also, \( \lim_{x \to 0} \cos(4x) = 1 \).

Therefore:

\[ = \frac{2}{3} \cdot \frac{1}{1} = \frac{2}{3} \]