mathop {lim }limits_{x to 1} frac{(2x - 3)(sq | vedclass

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Question

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3}$.First,factor the denominator: $2x^2

Vedclass · Accepted Answer

$-1/10$

Vedclass · Answer

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3}$.First,factor the denominator: $2x^2 + x - 3 = (2x + 3)(x - 1)$.Substitute this into the limit: $\mathop {\lim }\limits_{x 	o 1} \frac{(2x - 3)(\sqrt{x} - 1)}{(2x + 3)(x - 1)}$.Since $(x - 1) = (\sqrt{x} - 1)(\sqrt{x} + 1)$,the expression becomes: $\mathop {\lim }\limits_{x 	o 1} \frac{(2x - 3)(\sqrt{x} - 1)}{(2x + 3)(\sqrt{x} - 1)(\sqrt{x} + 1)}$.Cancel the common factor $(\sqrt{x} - 1)$: $\mathop {\lim }\limits_{x 	o 1} \frac{2x - 3}{(2x + 3)(\sqrt{x} + 1)}$.Now,substitute $x = 1$: $\frac{2(1) - 3}{(2(1) + 3)(\sqrt{1} + 1)} = \frac{-1}{(5)(2)} = -\frac{1}{10}$.

$\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3} = $

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