Limit of (x^n-1)/(x-1) as x approaches 1

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

lim_x→a (xⁿ-aⁿ)/(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: limx→1 (xn-1)/(x-1) is equal to n.

Explanation:

First, we apply the binomial formula:

xⁿ-1 = (x-1) (x^n-1+x^n-2+…+x+1).

Thus, the given limit

lim_x→1 (xⁿ-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 (xⁿ-1)/(x-1) is equal to n when x tends to 1.

Alternative Method: The limit of (xⁿ-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 (xⁿ-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 (xⁿ-1)/(x-1) is n when x goes to 1, i.e., lim_x→1 (xⁿ-1)/(x-1) = n.

Question-Answer

Question: Find lim_x→1 (x²ⁿ-1)/(x-1)

Answer:

limx→1 (x2n-1)/(x-1) = limx→1 $\dfrac{\frac{d}{dx}(x^n-1)}{\frac{d}{dx}(x-1)}$ = limx→1 $\dfrac{2nx^{2n-1}}{1}$ ⇒ limx→1 (x2n-1)/(x-1) = $2n \times 1^{2n-1}$ ⇒ limx→1 (x2n-1)/(x-1) = 2n. Thus the limit of (x2n-1)/(x-1) when x tends to 1 is equal to 2n.

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