Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 3

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PROBLEM:

Evaluate limx(4x+5x2+1)\displaystyle \lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\right)


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Solution:

Divide by the highest denominator power

limx(4x+5x2+1)=limx(4x+5x2+11x21x2)=limx(4xx2+5x2x2x2+1x2)=limx(4x+5x21+1x2)=0+01+0=0  (Answer)\begin{align*} \lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\right) & =\lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\cdot \frac{\displaystyle\frac{1}{x^2}}{\displaystyle\frac{1}{x^2}}\right) \\ \\ &=\lim\limits_{x\to \infty }\left(\frac{\displaystyle\frac{4x}{x^2}+\displaystyle\frac{5}{x^2}}{\displaystyle\frac{x^2}{x^2}+\displaystyle\frac{1}{x^2}}\right)\\ \\ &=\lim\limits_{x\to \infty }\left(\frac{\displaystyle\frac{4}{x}+\displaystyle\frac{5}{x^2}}{1+\displaystyle\frac{1}{x^2}}\right) \\ \\ &=\displaystyle\frac{0+0}{1+0} \\ \\ &=0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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