Now let there be a function be f(x) and let us assume that you are finding its left and right limits as x approaches a particular value, say 'a'.

### Left limit

In order to find the left hand limit of f(x) as x approaches 'a', or `lim_(x -> a^-)f(x)`, we start from a value just to the left of x = a on the graph of f(x) and move in closer and closer to x = a. It is clear that as this distance gets smaller and smaller the left limit becomes more and more accurate. Thus, if we start at a distance 'h' from x = a towards its left on the x-axis, then as 'h' gets smaller and smaller the left limit gets more and more accurate. At each smaller and smaller 'h' value, the function value at that point is f(a - h). Now as h gets smaller and smaller we area approaching x = a from the left side, so the left limit can be written as `lim_(h -> 0) f(a - h)`.Thus in order to find the left hand limit of f(x) as x approaches 'a', you will evaluate the following limit:

`lim_(x -> a^-) f(x) = lim_(h -> 0) f(a - h)`

### Right limit

Similarly, in order to find the right hand limit of f(x) as x approaches 'a', that is `lim_(x -> a^+) f(x)` you will start at a value just to the right of x = a and get in closer and closer to x = a. Let's say you start at a distance of 'h' units to the right of x = a, then as h becomes smaller and smaller you get a more and more accurate right limit of f(x) as x approaches x = a. Then this limit is given by `lim_(h -> 0) f(a + h)` because the function's value at a distance of 'h' from x = a towards its right is given by f(a + h) and the right limit of f(x) as x approaches 'a' is given by the limit of f(a + h) as h becomes smaller and smaller, that is it approaches 0.Thus in order to find the right hand limit of f(x) as x approaches 'a', you will evaluate the following limit:

`lim_(x -> a^+) f(x) = lim_(h -> 0) f(a + h)`

### Example:

Find the left and right limits of f(x) = x^2 + 2x + 1 as x approaches 10.### Solution:

#### Left hand limit

Left hand limit of f(x) as x approaches a is given by,

`lim_(h -> 0) f(a - h)`In the above formula, 'h' always remains as it is for all types of functions, because it is the variable approaching zero (the distance from x = a from just left of it). On the other hand, 'a' is the value of x of which the left hand limit is being calculated. Thus 'a' is 10,

`lim_(h -> 0) f(10 - h)`Now evaluate the function at x = 10 - h,

`lim_(h -> 0) (10 - h)^2 + 2(10 - h) + 1`Now substitute h = 0 in order to find the limit,

`(10 - 0)^2 + 2(10 - 0) + 1 = 100 + 20 + 1 = 121`Hence, the left hand limit of f(x) = x^2 + 2x + 1 as x approaches 10 is,

`lim_(x -> 10^-) f(x) = 121`

#### Right hand limit

The right hand limit evaluation is same as that for the left limit, using the right limit formula,

`lim_(h -> 0) f(a + h)`... where 'a' is 10,

`lim_(h -> 0) f(10 + h)`Evaluate the function at x = 10 + h,

`lim_(h -> 0) (10 + h)^2 + 2(10 + h) + 1`Substituting h = 0 to find the right limit,

`(10 + 0)^2 + 2(10 + 0) + 1 = 100 + 20 + 1 = 121`Thus the right limit of f(x) as x approaches 10 is 121.

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