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

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Question

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{2{x^2} - 3x + 1}}{{{x^2} - 1}}$,we divide the numerator and the denominator b

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{2{x^2} - 3x + 1}}{{{x^2} - 1}}$,we divide the numerator and the denominator by the highest power of $x$,which is ${x^2}$.$\mathop {\lim }\limits_{x 	o \infty } \frac{{2 - (3/x) + (1/{x^2})}}{{1 - (1/{x^2})}}$As $x 	o \infty$,the terms $(3/x)$,$(1/{x^2})$,and $(1/{x^2})$ approach $0$.Therefore,the limit becomes $\frac{{2 - 0 + 0}}{{1 - 0}} = \frac{2}{1} = 2$.

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

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