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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 111 of 150

Marks: +1, -0

\lim _{x \rightarrow 1}\left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]=

Solution:

Concept:

Factorize the denominators and simplify the algebraic expression before applying the limit.

Explanation:

The given limit is

\lim_{x \to 1} \left[ \frac{x-2}{x^2-x} - \frac{1}{x^3-3x^2+2x} \right]

First, factorize the denominators:

x^2 - x = x(x-1)

x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x-1)(x-2)

Rewrite the expression using the common denominator

x(x-1)(x-2)

\frac{x-2}{x(x-1)} - \frac{1}{x(x-1)(x-2)} = \frac{(x-2)^2 - 1}{x(x-1)(x-2)}

Simplify the numerator:

(x-2)^2 - 1 = x^2 - 4x + 4 - 1 = x^2 - 4x + 3 = (x-1)(x-3)

Thus, the expression becomes:

\frac{(x-1)(x-3)}{x(x-1)(x-2)}

Cancel the common factor

(x-1)

\frac{x-3}{x(x-2)}

Now substitute

x = 1

\frac{1-3}{1(1-2)} = \frac{-2}{-1} = 2

Therefore, the limit is

2

Answer:

2

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