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

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(A) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty} \frac{2x^2 + 3x + 4}{3x^2 + 3x + 4}$,divide the numerator and the denominator by the

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty} \frac{2x^2 + 3x + 4}{3x^2 + 3x + 4}$,divide the numerator and the denominator by the highest power of $x$,which is $x^2$.$\mathop {\lim }\limits_{x 	o \infty} \frac{2 + \frac{3}{x} + \frac{4}{x^2}}{3 + \frac{3}{x} + \frac{4}{x^2}}$As $x 	o \infty$,the terms $\frac{3}{x}$,$\frac{4}{x^2}$,$\frac{3}{x}$,and $\frac{4}{x^2}$ all approach $0$.Therefore,the limit is $\frac{2 + 0 + 0}{3 + 0 + 0} = \frac{2}{3}$.

$\mathop {\lim }\limits_{x \to \infty} \frac{2x^2 + 3x + 4}{3x^2 + 3x + 4}$ is equal to

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