mathop {lim }limits_{x to 1} frac{{{x^3} - 1} | vedclass

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Question

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 1} \frac{{{x^3} - 1}}{{{x^2} + 5x - 6}}$,we first check the form by substituting $x = 1$.Subs

Vedclass · Accepted Answer

$\frac{3}{7}$

Vedclass · Answer

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 1} \frac{{{x^3} - 1}}{{{x^2} + 5x - 6}}$,we first check the form by substituting $x = 1$.Substituting $x = 1$ gives $\frac{1^3 - 1}{1^2 + 5(1) - 6} = \frac{0}{0}$,which is an indeterminate form.We factorize the numerator and the denominator:Numerator: $x^3 - 1 = (x - 1)(x^2 + x + 1)$Denominator: $x^2 + 5x - 6 = (x - 1)(x + 6)$Now,the limit becomes $\mathop {\lim }\limits_{x 	o 1} \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 6)}$.Canceling the common factor $(x - 1)$,we get $\mathop {\lim }\limits_{x 	o 1} \frac{x^2 + x + 1}{x + 6}$.Substituting $x = 1$ gives $\frac{1^2 + 1 + 1}{1 + 6} = \frac{3}{7}$.

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{{x^2} + 5x - 6}} = $

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