Limit at Infinity| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.4, Problem 2|

Evaluate

\lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\right)

SOLUTION:

Divide by the highest denominator power 

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

=\lim\limits_{x\to \infty }\left(\frac{\frac{3x^2}{x^3}+\frac{x}{x^3}+\frac{2}{x^3}}{\frac{x^3}{x^3}+\frac{8x}{x^3}+\frac{1}{x^3}}\right)

=\lim\limits_{x\to \infty }\left(\frac{\frac{3}{x}+\frac{1}{x^2}+\frac{2}{x^3}}{1+\frac{8}{x^2}+\frac{1}{x^3}}\right)

=\frac{0+0+0}{1+0+0}

=0

 

 

 

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