mathop {lim }limits_{x to 1} frac{x - 1}{2x^2 | vedclass

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Question

(C) Given limit: $\mathop {\lim }\limits_{x 	o 1} \frac{x - 1}{2x^2 - 7x + 5}$.Substituting $x = 1$,we get the indeterminate form $\frac{0}{0}$.Facto

Vedclass · Accepted Answer

$-1/3$

Vedclass · Answer

(C) Given limit: $\mathop {\lim }\limits_{x 	o 1} \frac{x - 1}{2x^2 - 7x + 5}$.Substituting $x = 1$,we get the indeterminate form $\frac{0}{0}$.Factorizing the denominator: $2x^2 - 7x + 5 = (x - 1)(2x - 5)$.Thus,$\mathop {\lim }\limits_{x 	o 1} \frac{x - 1}{(x - 1)(2x - 5)} = \mathop {\lim }\limits_{x 	o 1} \frac{1}{2x - 5}$.Substituting $x = 1$: $\frac{1}{2(1) - 5} = \frac{1}{-3} = -\frac{1}{3}$.Alternatively,using $L$-Hospital's rule: $\mathop {\lim }\limits_{x 	o 1} \frac{\frac{d}{dx}(x - 1)}{\frac{d}{dx}(2x^2 - 7x + 5)} = \mathop {\lim }\limits_{x 	o 1} \frac{1}{4x - 7} = \frac{1}{4(1) - 7} = -\frac{1}{3}$.

$\mathop {\lim }\limits_{x \to 1} \frac{x - 1}{2x^2 - 7x + 5} = $

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