# Reciprocal Function f(x) = 1/x: Graph, Derivative, Integral and Limits

> The reciprocal function 1/x: its hyperbola graph and asymptotes, the derivative -1/x², why its integral is ln|x| + C, and its one-sided limits.

Source: https://practicecalculus.com/functions/reciprocal-function
Updated: 2026-11-03

$$ f(x) = \frac{1}{x} $$

## At a glance

- **Domain:** $x \ne 0$
- **Range:** $y \ne 0$
- **Derivative:** $-\frac{1}{x^2}$
- **Integral:** $\ln|x| + C$
- **Asymptotes:** $x = 0$ and $y = 0$
- **Symmetry:** Odd: $f(-x) = -f(x)$

## Key takeaways

- The reciprocal function $f(x) = \frac{1}{x}$ is defined for every $x$ except $0$ and its graph is a hyperbola.
- Its derivative is $-\frac{1}{x^2}$, negative everywhere, so it decreases on each piece of its domain.
- Its integral is $\ln|x| + C$, the one case the reverse power rule cannot handle.
- The one-sided limits at $0$ are $+\infty$ and $-\infty$, so $\lim_{x \to 0}\frac{1}{x}$ does not exist.

The **reciprocal function** $f(x) = \frac{1}{x}$ sends each number to one over itself. It is the simplest [rational function](/functions/rational-functions), the power function with exponent $-1$, and the standard example of infinite limits, asymptotes and the natural logarithm's derivative.

## Formula

$$f(x) = \frac{1}{x} = x^{-1}, \qquad x \ne 0$$

## Graph of 1/x

The graph is a **hyperbola** with two branches, one in the first quadrant and one in the third.

```graph
title: 'Graph of the reciprocal function y = 1/x'
caption: 'The graph approaches the vertical asymptote $x = 0$ and the horizontal asymptote $y = 0$ without touching either.'
x: [-5, 5]
y: [-5, 5]
functions:
  - expr: 1/x
    label: y = 1/x
points:
  - [1, 1, '(1, 1)']
  - [-1, -1, '(-1, -1)']
vertical:
  - {x: 0, label: 'x = 0'}
```

## Key properties

- **Domain and range.** All reals except $0$, for both.
- **Odd function.** $\frac{1}{-x} = -\frac{1}{x}$, so the graph is symmetric about the origin.
- **Asymptotes.** Vertical asymptote $x = 0$, horizontal asymptote $y = 0$.
- **Its own inverse.** Solving $y = \frac{1}{x}$ for $x$ gives $x = \frac{1}{y}$, so $f^{-1} = f$ and the graph is symmetric about $y = x$.
- **Discontinuous only at $0$,** which is not in the domain. It is continuous on its domain.

## Limits of 1/x

$$\lim_{x \to 0^+}\frac{1}{x} = +\infty, \qquad \lim_{x \to 0^-}\frac{1}{x} = -\infty, \qquad \lim_{x \to \pm\infty}\frac{1}{x} = 0$$

Because the one-sided limits at $0$ disagree, $\lim_{x \to 0}\frac{1}{x}$ does not exist. The limit at infinity is the basic building block for [limits at infinity](/blog/limits-at-infinity) of every rational function.

## Derivative of 1/x

$$\frac{d}{dx}\,\frac{1}{x} = -\frac{1}{x^2}$$

This follows from the [power rule](/rules/power-rule) with $n = -1$. The derivative is negative everywhere it is defined, so the function decreases on each branch, and the tangent lines get steeper near $0$ and flatter far away. The second derivative is $\frac{2}{x^3}$: concave up for $x > 0$, concave down for $x < 0$.

## Integral of 1/x

$$\int \frac{1}{x}\,dx = \ln|x| + C$$

The reverse power rule would give $\frac{x^0}{0}$, which is undefined, so $\frac{1}{x}$ is the one power that integrates to a [natural logarithm](/functions/natural-logarithm). The absolute value covers negative $x$: for $x < 0$, $\frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac{1}{x}$.

A definite integral cannot cross $x = 0$: $\int_{-1}^{1}\frac{1}{x}\,dx$ is an [improper integral](/blog/improper-integrals) that diverges.

## Worked examples

**Example 1: tangent line.** At $x = 2$, $f(2) = \frac{1}{2}$ and $f'(2) = -\frac{1}{4}$, so the tangent line is $y = \frac{1}{2} - \frac{1}{4}(x - 2) = -\frac{1}{4}x + 1$.

**Example 2: definite integral.** $\int_1^e \frac{1}{x}\,dx = \ln e - \ln 1 = 1$.

**Example 3: a shifted reciprocal.** $\frac{d}{dx}\frac{3}{x - 1} = -\frac{3}{(x - 1)^2}$ and $\int \frac{3}{x - 1}\,dx = 3\ln|x - 1| + C$.

## Common mistakes

1. **Writing $\int \frac{1}{x}\,dx = \frac{x^0}{0}$** or $\ln x$ without the absolute value.
2. **Getting the derivative sign wrong.** $\frac{d}{dx}\frac{1}{x}$ is $-\frac{1}{x^2}$, not $\frac{1}{x^2}$.
3. **Confusing $\frac{1}{x}$ with $x^{-1}$ as an inverse function.** $\frac{1}{f(x)}$ is a reciprocal, while $f^{-1}(x)$ usually means an inverse. For $\frac{1}{x}$ they happen to agree, which is unusual.
4. **Integrating across the asymptote** as if the function were continuous there.

## How it connects

The area under $\frac{1}{t}$ from $1$ to $x$ is exactly $\ln x$, which is one way to define the natural logarithm. The [power functions](/functions/power-functions) page covers the family $x^n$ that $\frac{1}{x}$ belongs to, and the harmonic series $\sum \frac{1}{n}$ is its discrete cousin.

## Frequently asked questions

### What is the derivative of 1/x?

Write $\frac{1}{x} = x^{-1}$ and use the power rule: $\frac{d}{dx}x^{-1} = -x^{-2} = -\frac{1}{x^2}$.

### What is the integral of 1/x?

The integral is $\int \frac{1}{x}\,dx = \ln|x| + C$. The reverse power rule fails here because it would divide by $n + 1 = 0$. The absolute value makes the formula valid for negative $x$ too.

### What is the limit of 1/x as x approaches 0?

The limit does not exist. From the right $\frac{1}{x} \to +\infty$ and from the left $\frac{1}{x} \to -\infty$, so the two one-sided limits disagree.

### Is 1/x decreasing?

It decreases on $(-\infty, 0)$ and on $(0, \infty)$ separately, since its derivative is negative. It is not decreasing on its whole domain, because $\frac{1}{-1} = -1 < 1 = \frac{1}{1}$.