The sum rule of limits says that the limit of the sum of two functions is the same as the sum of the limits of the individual functions. In this post, we will prove the sum/addition rule of limits by the epsilon-delta method.

**Question:**What is an epsilon-delta proof of the sum law of limits? Here, you will find the answer given below.

## Sum law of limits

If the limits of two functions f and g exist when x tends to a, then

`lim_{x to a}(f+g)` `=lim_{x to a} f + lim_{x to a} g`

**Proof:**

To prove the sum law of limits, we will use the epsilon-delta method. Let `lim_{x to a} f(x) = A` and `lim_{x to a} g(x) = B`. So by the epsilon-delta definition of limits, we have the following:

For a given `epsilon>0`, there exists `delta_1>0` and `delta_2>0` such that

`|f(x)-A| < epsilon/2`, whenever `0< |x-a| < delta_1`

**…(I)**

and

`|g(x)-B| < epsilon/2`, whenever `0< |x-a| < delta_2`

**…(II)**Set `delta: = text{min} {delta_1, delta_2 }`.

Then whenever `0< |x-a| < delta`, we have that

`|f(x)+g(x) – (A+B)|` `= |(f(x) – A) + (g(x) – B)|`

`leq |f(x) – A| + |g(x) – B|` by the triangle inequality |a+b| ≤ |a|+|b|

`leq epsilon/2 + epsilon/2` by

**(I)**and**(II)**

`=epsilon`

Thus, `lim_{x to a}(f+g)` `=lim_{x to a} f + lim_{x to a} g` and we have proved the sume rule / addition rule of limits.

## Examples

We know that `lim_{xto 0} x^2=0` and `lim_{xto 0} 1=1`.

So by the above sum rule of limits, the limit of `x^2+1` when x tends to 0 is equal to

`lim_{xto 0} (x^2+1)`

= `lim_{xto 0} x^2+ lim_{xto 0} 1`

= `0+1`

= `1`