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AP ICET 2024 6th May 2024 Shift 1

Question : 170 of 200

Marks: +1, -0

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

Solution:

We are given:

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

Factor the denominator of the second fraction:

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

Thus, the expression becomes:

\lim\limits_{x \to 2} \left[ \frac{1}{x-2} - \frac{1}{(x-1)(x-2)} \right] .

Now, find a common denominator:

= \lim\limits_{x \to 2} \frac{(x-1)-1}{(x-2)(x-1)} = \lim\limits_{x \to 2} \frac{x-2}{(x-2)(x-1)}

Canceling

(x-2)

from the numerator and denominator, we get:

= \lim\limits_{x \to 2} \frac{1}{x-1}

Substitute

x=2

= \frac{1}{2-1} = 1 .

Quick Tip

When simplifying limits with common terms in the numerator and denominator,

factor the expression and cancel out common factors.

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