W wendywoo New member Joined Jun 12, 2011 Messages 31 Jun 19, 2011 #1 Evaluate lim (x/??? x^2-2) as x goes into infinity. I don't remember what to do if the bottom is a radical.
Evaluate lim (x/??? x^2-2) as x goes into infinity. I don't remember what to do if the bottom is a radical.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Jun 19, 2011 #2 Since \(\displaystyle x\to \infty\), get rid of the 2 and write as: \(\displaystyle \lim_{x\to \infty}\frac{x}{\sqrt{x^{2}}}\) \(\displaystyle \lim_{x\to \infty}\frac{x}{|x|}\) Since we are heading in the positive direction toward infinity: \(\displaystyle \lim_{x\to \infty}\frac{x}{x}=1\) That's all it is. You can do this when \(\displaystyle x\to \pm\infty\)
Since \(\displaystyle x\to \infty\), get rid of the 2 and write as: \(\displaystyle \lim_{x\to \infty}\frac{x}{\sqrt{x^{2}}}\) \(\displaystyle \lim_{x\to \infty}\frac{x}{|x|}\) Since we are heading in the positive direction toward infinity: \(\displaystyle \lim_{x\to \infty}\frac{x}{x}=1\) That's all it is. You can do this when \(\displaystyle x\to \pm\infty\)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jun 20, 2011 #3 Hello, wendywoo! Another approach . . . \(\displaystyle \text{Evaluate: }\;\lim_{x\to\infty}\frac{x}{\sqrt{ x^2-2}}\) Click to expand... Divide numerator and denominator by \(\displaystyle x\) . . \(\displaystyle \lim_{x\to\infty}\frac{\frac{x}{x}}{\frac{\sqrt{x^2-2}}{x}} \;=\;\lim_{x\to\infty}\frac{1}{\sqrt{\frac{x^2-2}{x^2}}} \;=\;\lim_{x\to\infty}\frac{1}{\sqrt{1-\frac{2}{x^2}}} \;=\;\frac{1}{\sqrt{1-0}} \;=\;1\)
Hello, wendywoo! Another approach . . . \(\displaystyle \text{Evaluate: }\;\lim_{x\to\infty}\frac{x}{\sqrt{ x^2-2}}\) Click to expand... Divide numerator and denominator by \(\displaystyle x\) . . \(\displaystyle \lim_{x\to\infty}\frac{\frac{x}{x}}{\frac{\sqrt{x^2-2}}{x}} \;=\;\lim_{x\to\infty}\frac{1}{\sqrt{\frac{x^2-2}{x^2}}} \;=\;\lim_{x\to\infty}\frac{1}{\sqrt{1-\frac{2}{x^2}}} \;=\;\frac{1}{\sqrt{1-0}} \;=\;1\)
