Limit at Infinity| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.4, Problem 1|

Evaluate

\lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\right)

SOLUTION:

Divide by the highest denominator power 

\lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\right)=\lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}}\right)

=\lim\limits_{x\to \infty \:}\left(\frac{6+\frac{4}{x}+\frac{5}{x^3}}{8+\frac{7}{x^2}-\frac{3}{x^3}}\right)

=\frac{\lim\limits_{x\to \infty \:}\left(6+\frac{4}{x}+\frac{5}{x^3}\right)}{\lim\limits_{x\to \infty \:}\left(8+\frac{7}{x^2}-\frac{3}{x^3}\right)}

=\frac{6+0+0}{8+0-0}

=\frac{6}{8}

=\frac{3}{4}

 

 

 

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