# Limit of (x^n-a^n)/(x-a) as x approaches a: Formula, Proof

The limit of (x^n-a^n)/(x-a) as x approaches a is equal to nan-1. This limit is denoted by limx→a (xn-an)/(x-a), so the limit formula of (xn-an)/(x-a) when x tends to a is given as follows.

Lets prove this limit formula.

## Proof of limx→a (xn-an)/(x-a)

To prove limx→a (xn-an)/(x-a) = n⋅an-1 we will consider three different cases depending upon n is a positive integer, negative integer, or a rational number.

Case 1:

n is a positive integer.

Using the binomial identity: xn-an = (x-a) (xn-1 +xn-2a +…+an-1) we get that

limx→a (xn-an)/(x-a)

= limx→a (xn-1 +xn-2a +…+an-1)

= an-1 +an-2a +…+an-1

= an-1 +an-1 +…+an-1 {n times}

= nan-1.

So the limit of (xn-an)/(x-a) is equal to nan-1 when x tends to a.

Case 2:

n is a negative integer.

Assume n= -m where m is a positive integer.

So the limit formula limx→a (xn-an)/(x-a) = nan-1 is valid when n is a negative integer.

Case 3:

n is any real number.

Assume n= p/q where q≠1 is a positive integer and p is either a positive or a negative integer.

Let x1/q = z and a1/q =b.

Therefore, x=zq and a=bq.

Observe, z→b when x→a.

Now,

$\dfrac{x^n-a^n}{x-a}$ $=\dfrac{x^{p/q}-a^{p/q}}{x-a}$ $=\dfrac{z^p-b^p}{z^q-b^q}$ $=\dfrac{\frac{z^p-b^p}{z-b}} {\frac{z^q-b^q}{z-b} }$

Therefore,

limx→a (xn-an)/(x-a)

= limz→b [(zp-bp)/(z-b) / (zq-bq)/(z-b)]

= limz→b (zp-bp)/(z-b) / limz→b (zq-bq)/(z-b)

= pbp-1 /qbq-1

= $\dfrac{p}{q}b^{p-q}$

= $\dfrac{p}{q}a^{\frac{p-q}{q}}$ since a1/q =b.

= $\dfrac{p}{q}a^{\frac{p}{q}-1}$

= nan-1

So the limit of (xn-an)/(x-a) is equal to nan-1 as x approaches a, and this is proved for any real number n. This completes the proof.

More Limits:

Limit of (xn-1)/(x-1) when x→1

Limit of x1/x when x→∞

Limit of xsin(1/x) when x→0

Limit of x2sin(1/x) when x→0

Limit of 1/x2 when x→∞

## FAQs

### Q1: What is the limit of (xn-an)/(x-a) when x tends to a?

Answer: The limit of (xn-an)/(x-a) is equal to nan-1 when x tends to a, that is, limx→a (xn-an)/(x-a) = nan-1.