limit question

[dugan]

1) lim x-00 x^2sin(1/x)

2) lim x-00 (x^2+3x)^(1/2)-x

 

[tkhunny]

1) lim x-00 x^2sin(1/x)

Really big clue:

$\displaystyle x^{2}\,*\,sin(\frac{1}{x})\,=\,x\,*\,\frac{sin(\frac{1}{x})}{\frac{1}{x}}$

 

[soroban]

Hello, dugan!

$\displaystyle \displaystyle{2)\;\;\lim_{x\to\infty}\left[\sqrt{x^2 + 3x}\ -\ x\right] }$

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: .$\displaystyle \displaystyle{\lim_{x\to\infty}\left(\frac{3}{\sqrt{1 + \frac{3}{x}}\ +\ 1}\right) \;= \;\frac{3}{\sqrt{1+0}\ +\ 1} \;= \;\frac{3}{2}}$