If lim _{x rightarrow infty}left(frac{11 x^3- | vedclass

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(D) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}$,we divide the numerator and the denominator by the highes

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(D) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}$,we divide the numerator and the denominator by the highest power of $x$,which is $x^3$: $\lim _{x \rightarrow \infty} \frac{x^3(11 - \frac{3}{x^2} + \frac{4}{x^3})}{x^3(13 - \frac{5}{x} - \frac{7}{x^3})}$ $\lim _{x \rightarrow \infty} \frac{11 - \frac{3}{x^2} + \frac{4}{x^3}}{13 - \frac{5}{x} - \frac{7}{x^3}}$ As $x \rightarrow \infty$,the terms $\frac{3}{x^2}, \frac{4}{x^3}, \frac{5}{x},$ and $\frac{7}{x^3}$ all approach $0$. Therefore,the limit is $\frac{11 - 0 + 0}{13 - 0 - 0} = \frac{11}{13}$. Comparing this to $\frac{a}{b}$,we get $a = 11$ and $b = 13$. Thus,$a + b = 11 + 13 = 24$.

If $\lim _{x \rightarrow \infty}\left(\frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}\right)=\frac{a}{b}$,then the value of $a+b$ equals:

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