Q. \[ \frac{\sin x}{x} \]

Q: What is \lim_{x \to 0} \frac{\sin x}{x}?

\lim_{x \to 0} \frac{\sin x}{x} = 1. Use the standard squeeze theorem (or the Taylor expansion \sin x = x - \frac{x^3}{6} + \cdots).

Q: Is \frac{\sin x}{x} defined at x=0?

It is undefined as written, but the limit exists. Define \frac{\sin x}{x} = 1 at x=0 to make it continuous.

Q: How do I compute \frac{d}{dx}\left(\frac{\sin x}{x}\right) for x \neq 0?

Use the quotient rule: \left(\frac{\sin x}{x}\right)' = \frac{x\cos x - \sin x}{x^2}, valid for x \neq 0.

Q: What is \lim_{x \to \infty} \frac{\sin x}{x}?

Since |\sin x| \le 1, \left|\frac{\sin x}{x}\right| \le \frac{1}{x}. Thus \lim_{x \to \infty} \frac{\sin x}{x} = 0.

Q: Solve \frac{\sin x}{x} = 0.

The fraction equals 0 when \sin x = 0 and x \neq 0. So x = n\pi for integers n \neq 0.

Q: Does \frac{\sin x}{x} have a removable discontinuity at x=0?

Yes. The limit is 1, so the discontinuity can be removed by defining the function value at 0 as 1.

Answer

\( \frac{\sin x}{x} \) is defined for \(x \ne 0\). At \(x=0\), use the limit:

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

So, \( \frac{\sin x}{x} = 1\) when \(x=0\), and it equals \( \frac{\sin x}{x}\) for \(x \ne 0\).

Detailed Explanation

We want to interpret and evaluate the expression

\[
\frac{\sin x}{x}.
\]

As written, \(\frac{\sin x}{x}\) is defined for all real \(x\) such that \(x \ne 0\). The only potential issue is at \(x=0\), because division by zero is not allowed.

Step 1: Check the domain (when \(x \ne 0\)).

If \(x \ne 0\), then \(\frac{\sin x}{x}\) is just a normal expression and is already defined.

Step 2: Discuss what happens at \(x=0\).

To meaningfully talk about the value at \(x=0\), we consider the limit:

\[
\lim_{x \to 0} \frac{\sin x}{x}.
\]

Step 3: Use a standard trigonometric limit.

A well-known limit in calculus is:

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

This holds when \(x\) is measured in radians.

Step 4: State the result.

So:

\[
\frac{\sin x}{x} \text{ is defined for } x \ne 0,
\]

and at \(x=0\) the limiting value is:

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

See full solution

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Calculus FAQ

What is \(\lim_{x \to 0} \frac{\sin x}{x}\)?

\(\lim_{x \to 0} \frac{\sin x}{x} = 1\). Use the standard squeeze theorem (or the Taylor expansion \(\sin x = x - \frac{x^3}{6} + \cdots\)).

Is \(\frac{\sin x}{x}\) defined at \(x=0\)?

It is undefined as written, but the limit exists. Define \(\frac{\sin x}{x} = 1\) at \(x=0\) to make it continuous.

How do I compute \(\frac{d}{dx}\left(\frac{\sin x}{x}\right)\) for \(x \neq 0\)?

Use the quotient rule: \(\left(\frac{\sin x}{x}\right)' = \frac{x\cos x - \sin x}{x^2}\), valid for \(x \neq 0\).

What is the series expansion of \(\frac{\sin x}{x}\) near \(x=0\)?

\(\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \cdots\). Comes from dividing the Taylor series of \(\sin x\) by \(x\).

What is \(\lim_{x \to \infty} \frac{\sin x}{x}\)?

Since \(|\sin x| \le 1\), \(\left|\frac{\sin x}{x}\right| \le \frac{1}{x}\). Thus \(\lim_{x \to \infty} \frac{\sin x}{x} = 0\).

Solve \(\frac{\sin x}{x} = 0\).

The fraction equals \(0\) when \(\sin x = 0\) and \(x \neq 0\). So \(x = n\pi\) for integers \(n \neq 0\).

Does \(\frac{\sin x}{x}\) have a removable discontinuity at \(x=0\)?

Yes. The limit is \(1\), so the discontinuity can be removed by defining the function value at \(0\) as \(1\).

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