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**Hint:**In this problem, we have to simplify this expression by using limits. If we assume that f(y) is a function, then the limit of the function can be represented as: $\displaystyle \lim_{y \to c}$. This is denoted as the general expression of limit, where c is any constant value. We will use a specific formula for limit in this question:

$\Rightarrow \displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$

**Complete step by step answer:**

Now, let’s solve the question.

As we know that the limit of a function f(x) is defined as a value, where the function reaches as the limit reaches some value. Limits are used for defining integration, integral calculus and continuity of the function. If we assume that f(y) is a function, then the limit of the function can be represented as;

$\displaystyle \lim_{y \to c}$

This is denoted as the general expression of limit, where c is any constant value. for this question, we will use a formula for limits and exponents:

$\Rightarrow \displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$

Now, write the given limit:

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

Multiply its numerator and denominator such that:

$\Rightarrow \displaystyle \lim_{x \to 1}\dfrac{{{x}^{m}}-1}{x-1}\times \dfrac{x-1}{{{x}^{n}}-1}$

According to the formula $\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$, a = 1. So, we can write:

$\Rightarrow \displaystyle \lim_{x \to 1}\dfrac{{{x}^{m}}-{{\left( 1 \right)}^{m}}}{x-1}\times \dfrac{x-1}{{{x}^{n}}-{{\left( 1 \right)}^{n}}}$

By applying formula, we will get:

$\Rightarrow m{{.1}^{m-1}}\times \dfrac{1}{n{{.1}^{n-1}}}$

Now after simplification, we will get:

$\therefore \dfrac{m}{n}$

**So, the correct answer is “Option b”.**

**Note:**Students should note that once you apply formula for limits and exponents, the limits will be removed and then the expression will be simplified. You may notice that before applying formula, we changed the expression according to the formula to be applied because if we apply any algebraic identity, it won’t work.

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