# Why is y=1/x a continuous function?

Oct 15, 2017

$y = \frac{1}{x}$ is NOT a continuous function.

#### Explanation:

$y = \frac{1}{x}$ is NOT a continuous function.

This function has a point of discontinuity at $x = 0$. This is because we cannot have 1/0, so there becomes an asymptote.

Similarly, $y \ne 0$.

So this function is NOT continuous as it has asymptotes along the lines $x = 0$ and $y = 0$.

Oct 15, 2017

$y = \frac{1}{x}$ is continuous over its domain. As zero is not in the domain, you can say $\frac{1}{x}$ is a continuous function.

#### Explanation:

$y = \frac{1}{x}$

We first need to determine the domain of $y$

$y$ is defined for $\forall x \ne 0$

Hence, the domain of $y$ is: $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

Now, $y$ is continious over $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

Since $y = \frac{1}{x}$ is continuous over its domain. As zero is not in the domain, you can say $\frac{1}{x}$ is a continuous function.

Oct 15, 2017

See below.

#### Explanation:

You seem to be getting conflicting information on this. One of the reasons is because continuity is generally referring to a given point or interval. The function is said to be discontinuous at a point if the limit at that point dosen't exist. So for a function to be continuous over the domain, the limit of any point in the domain must exist. For the function $\frac{1}{x}$, the limit dosen't exist at zero, so the function is not continuous over its domain, but is continuous either side of this, as can be seen from its graph. These ideas can become complicated with some functions, but this is the general idea.

Oct 15, 2017

The domain under consideration determines the answer here. The function $y = f \left(x\right) = \frac{1}{x}$ is continuous on $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$ but it is not continuous on $\mathbb{R} = \left(- \infty , \infty\right)$.

#### Explanation:

The function $y = f \left(x\right) = \frac{1}{x}$ is continuous for all $x$ in its "natural" domain, which is $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$. It's not even defined at $x = 0$, so it is not continuous on $\mathbb{R} = \left(- \infty , \infty\right)$. Even if we defined it to have a value at $x = 0$, it still would not be continuous on $\mathbb{R}$ because its discontinuity at $x = 0$ is not "removable" (in this case, the vertical asymptotes as $x$ approaches 0 from either the right or left are the reasons the discontinuity at 0 is not removable).

On the other hand, the function $f \left(x\right) = \sin \frac{x}{x}$ has a discontinuity at $x = 0$ that is removable. The fact that ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$ implies that we can define $f$ in a piecewise way so that $f \left(0\right) = 1$ in order to make a continuous function on $\mathbb{R}$.