In our study of limits of a function `Lim_(x→a) u(X)`, we defined `x` to exist in an open interval containing a. This means that we assumed that `u(X)` is defined on both sides of a. However, in calculus we also study and evaluate limits w.r.t. one side only(i.e., either for values of x greater than a or those less than a). Such limits are known as One-sided limits.

As evident from the name, one-sided limit can be of **two types**:

`Lim_(x→a-) u(X)` is defined to be the limit of `u(X)`as x approaches a from the left hand side, i.e., this limit is defined for values of x sufficiently close to a but less than a.

`Lim_(x→a+) u(X)` is defined to be the limit of `u(X)`as x approaches a from the right hand side, i.e., this limit is defined for values of x sufficiently
close to a but greater than a.

As regards the evaluation of one-sided limits, you do not need to be confused about them. All the theorems and solution techniques of limits discussed above
are equally applicable for the evaluation of one-sided limits.

We have gone through the concept of one-sided limits because they are pre-requisite for developing an understanding of a continuous / discontinuous function.
Continuity/discontinuity of a function is a topic that you will find frequently in your Mathematics courses, and having a good understanding on the topic will
be really worthwhile. We discuss it next.

There are three conditions that need to be met by a function f in order to be continuous at a number a. These are:

- `f(a)` is defined.
- `Lim_(x→a) f(x)` exists.
- `Lim_(x→a) f(x) = f(a)`

If any one or more of these three conditions is not true for `f ` at "a", then the function `f` is called a discontinuous function at "a".

In order to see whether the given function is continuous or discontinuous at x = –1, we will have to check the function for all the three
conditions specified above one by one.

We start with checking for the first condition (f(a) is defined).

We are given that

` f(x) = ( (x^2 – 1) )/(x + 1 )`

`f(–1) = ( ((–1)^2 – 1) )/(–1 + 1 )= ∞`

`f(–1) = ( ((–1)^2 – 1) )/(–1 + 1 )= ∞`

Hence f(–1) is not defined. We said above that if any of the three conditions of continuity is violated, function is said to be
discontinuous.

=>f(x) is discontinuous at –1.

However, if we try to find the Limit of `f(x)`, we conclude that `f(x)` is continuous on all the values other than –1.

`Lim_(x → -1 ) f(x)` = ` Lim_(x → -1) ( (x^2 – 1) )/(x + 1 ) ` = `Lim_(x → -1) ( x – 1 ) ` = `(– 1 – 1 ) = –2`

This implies that `f(x)` is continuous at all the values of x other than – 1.

We check for all the three conditions of continuity one by one.

**Checking for the first condition (f(a) is defined).**

`f(3) = 2(3) + 1 = 7 ` Hence the first condition is true.

**Checking for the second condition (`Lim_(x → a) f(x)` exists).**

`Lim_(x → 3-) f(x)` = `Lim_(x → 3-) (x – 1)` = 3 – 1 = 2

`Lim_(x → 3+) f(x)` = `Lim_(x → 3+) ( 2x + 1 )` = 2(3) + 1 = 7

∵ `Lim_(x → 3-) f(x)` ≠ `Lim_(x → 3+) f(x)`**i.e. (L.H. Limit ≠ R.H. Limit)**

`Lim_(x → 3+) f(x)` = `Lim_(x → 3+) ( 2x + 1 )` = 2(3) + 1 = 7

∵ `Lim_(x → 3-) f(x)` ≠ `Lim_(x → 3+) f(x)`

`Lim_(x → 3) f(x)` does not exist.

Hence `f(x)` is not continuous at x = 3.

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