$\lim_{x \to 1} \dfrac{x^m - 1}{x^n - 1}$ is (A) $1$ (B) $\dfrac{m}{n}$ (C) $-\dfrac{m}{n}$ (D) $\dfrac{m^2}{n^2}$

limx → 1 (xm - 1)/(xⁿ - 1) is (A) 1 (B) (m)/(n) (C – Class 11 Mathematics

NCERT Exemplar · Q57

\lim_{x \to 1} \dfrac{x^m - 1}{x^n - 1}

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Direct substitution gives $\frac{0}{0}$ . Dividing numerator and denominator by $(x-1)$ and using the standard limit $\lim_{x\to 1}\frac{x^k-1}{x-1}=k$ gives $\frac{m}{n}$ — option (B).

Step 1 — Check by substitution.

At $x = 1$ : numerator $1^m - 1 = 0$ , denominator $1^n - 1 = 0$ . So we have $\frac{0}{0}$ , indeterminate, and must simplify.

Step 2 — Recall the standard algebraic limit.

\lim_{x\to a}\dfrac{x^k - a^k}{x-a} = k\,a^{k-1}

Taking $a = 1$ (and $1^{k-1}=1$ ):

\lim_{x\to 1}\frac{x^k - 1}{x-1} = k.

Step 3 — Split the fraction using $(x-1)$ .

Both $x^m - 1$ and $x^n - 1$ vanish at $x=1$ , so each has a factor $(x-1)$ . Divide numerator and denominator by $(x-1)$ :

$\frac{x^m - 1}{x^n - 1} = \frac{\dfrac{x^m - 1}{x-1}}{\dfrac{x^n - 1}{x-1}}.$ …

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Q. $\lim_{x \to 1} \dfrac{x^m - 1}{x^n - 1}$ is
(A) $1$
(B) $\dfrac{m}{n}$
(C) $-\dfrac{m}{n}$
(D) $\dfrac{m^2}{n^2}$

Concept understanding — Standard Limit of (xⁿ − aⁿ)/(x − a)

Q.lim⁡x→1xm−1xn−1\lim_{x \to 1} \dfrac{x^m - 1}{x^n - 1}limx→1​xn−1xm−1​ is (A) 111 (B) mn\dfrac{m}{n}nm​ (C) −mn-\dfrac{m}{n}−nm​ (D) m2n2\dfrac{m^2}{n^2}n2m2​

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Q. $\lim_{x \to 1} \dfrac{x^m - 1}{x^n - 1}$ is
(A) $1$
(B) $\dfrac{m}{n}$
(C) $-\dfrac{m}{n}$
(D) $\dfrac{m^2}{n^2}$

Unlock everything free for 14 days