limit question
- Thread starter dugan
- Start date
Hello, dugan!
. . \(\displaystyle \displaystyle{\frac{\sqrt{x^2 + 3x}\ -\ x}{1}\ \cdot\ \frac{\sqrt{x^2 + 3x}\ +\ x}{\sqrt{x^2 + 3x}\ +\ x} \;= \;\frac{(x^2 + 3x)\ -\ x^2}{\sqrt{x^2+3x}\ +\ x} \;= \;\frac{3x}{\sqrt{x^2+3x}\ +\ x} }\)
Divide top and bottom by \(\displaystyle x:\)
. . \(\displaystyle \displaystyle{\frac{\frac{3x}{x}}{\frac{\sqrt{x^2+3x}}{x}\ +\ \frac{x}{x}} \;= \;\frac{3}{\sqrt{\frac{x^2}{x^2} + \frac{3x}{x^2}}\ +\ 1} \;= \;\frac{3}{\sqrt{1 + \frac{3}{x}}\ +\ 1} }\)
Take the limit: .\(\displaystyle \displaystyle{\lim_{x\to\infty}\left(\frac{3}{\sqrt{1 + \frac{3}{x}}\ +\ 1}\right) \;= \;\frac{3}{\sqrt{1+0}\ +\ 1} \;= \;\frac{3}{2}}\)
Multiply top and bottom by \(\displaystyle \sqrt{x^2 + 3x}\ +\ x:\)
. . \(\displaystyle \displaystyle{\frac{\sqrt{x^2 + 3x}\ -\ x}{1}\ \cdot\ \frac{\sqrt{x^2 + 3x}\ +\ x}{\sqrt{x^2 + 3x}\ +\ x} \;= \;\frac{(x^2 + 3x)\ -\ x^2}{\sqrt{x^2+3x}\ +\ x} \;= \;\frac{3x}{\sqrt{x^2+3x}\ +\ x} }\)
Divide top and bottom by \(\displaystyle x:\)
. . \(\displaystyle \displaystyle{\frac{\frac{3x}{x}}{\frac{\sqrt{x^2+3x}}{x}\ +\ \frac{x}{x}} \;= \;\frac{3}{\sqrt{\frac{x^2}{x^2} + \frac{3x}{x^2}}\ +\ 1} \;= \;\frac{3}{\sqrt{1 + \frac{3}{x}}\ +\ 1} }\)
Take the limit: .\(\displaystyle \displaystyle{\lim_{x\to\infty}\left(\frac{3}{\sqrt{1 + \frac{3}{x}}\ +\ 1}\right) \;= \;\frac{3}{\sqrt{1+0}\ +\ 1} \;= \;\frac{3}{2}}\)