The limit of (x^n-1)/(x-1) as x approaches 1 is equal to n, that is, lim_{x→1} (x^{n}-1)/(x-1) = n. This follows from the formula

lim_{x→a} (x^{n}-a^{n})/(x-a) = n⋅a^{n-1}

Put a=1, so we get that

lim_{x→1} $\dfrac{x^n-1}{x-1}$ = n.

Let us now prove this limit formula.

## What is the Limit of (x^n-1)/(x-1)

Answer: lim_{x→1} (x^{n}-1)/(x-1) is equal to n. |

**Explanation:**

First, we apply the binomial formula:

x^{n}-1 = (x-1) (x^{n-1}+x^{n-2}+…+x+1).

Thus, the given limit

lim_{x→1} (x^{n}-1)/(x-1)

= lim_{x→1} $\dfrac{(x-1)(x^{n-1}+\cdots+x+1)}{x-1}$

= lim_{x→1} (x^{n-1}+x^{n-2}+…+x+1)

= 1^{n-1}+1^{n-2}+…+1+1

= 1+1+…+1+1 {n times}

= n.

So the limit of (x^{n}-1)/(x-1) is equal to n when x tends to 1.

## Limit of (x^{n}-1)/(x-1) using L’Hôpital’s Rule

The limit of (x^{n}-1)/(x-1) when x→1 can be computed using the L’Hôpital’s rule as it is an intermediate form (0/0).

Therefore,

lim_{x→1} (x^{n}-1)/(x-1)

= lim_{x→1} $\dfrac{\frac{d}{dx}(x^n-1)}{\frac{d}{dx}(x-1)}$

= lim_{x→1} $\dfrac{nx^{n-1}}{1}$

= $n \times 1^{n-1}$

= n × 1

= n.

Hence we have shown that the limit of (x^{n}-1)/(x-1) is n when x goes to 1, i.e., lim_{x→1} (x^{n}-1)/(x-1) = n.

## Question-Answer

**Question:** Find lim_{x→1} (x^{2n}-1)/(x-1)

**Answer:**

lim_{x→1} (x^{2n}-1)/(x-1) = lim_{x→1} $\dfrac{\frac{d}{dx}(x^n-1)}{\frac{d}{dx}(x-1)}$ = lim_{x→1} $\dfrac{2nx^{2n-1}}{1}$⇒ lim _{x→1} (x^{2n}-1)/(x-1) = $2n \times 1^{2n-1}$⇒ lim _{x→1} (x^{2n}-1)/(x-1) = 2n.Thus the limit of (x ^{2n}-1)/(x-1) when x tends to 1 is equal to 2n. |

Have You Read These Limits?

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## FAQs

### Q1: What is lim_{x→1} (x^{n}-1)/(x-1)?

Answer: The value of the limit lim_{x→1} (x^{n}-1)/(x-1) is equal to n.