Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 5

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PROBLEM:

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


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SOLUTION:

Divide by the highest denominator power

\begin{align*}

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

& =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{8x}{x}-\displaystyle \frac{5}{x}}{\sqrt{\displaystyle \frac{4x^2}{x^2}+\displaystyle \frac{3}{x^2}}}\right)\\
\\
& =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8-\displaystyle \frac{5}{x}}{\displaystyle \sqrt{4+\displaystyle \frac{3}{x^2}}}\right) \\

\\
& =\displaystyle \frac{8-0}{\sqrt{4+0}} \\
\\
& =\displaystyle \frac{8}{2} \\
\\
& =\displaystyle 4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

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