evaluating limits

[wendywoo]

Evaluate lim (x/??? x^2-2) as x goes into infinity.

I don't remember what to do if the bottom is a radical.

 

[galactus]

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 \(\displaystyle x\to \pm\infty\)

 

[soroban]

Hello, wendywoo!

Another approach . . .

\(\displaystyle \text{Evaluate: }\;\lim_{x\to\infty}\frac{x}{\sqrt{ x^2-2}}\)

Divide numerator and denominator by \(\displaystyle x\)

. . \(\displaystyle \lim_{x\to\infty}\frac{\frac{x}{x}}{\frac{\sqrt{x^2-2}}{x}} \;=\;\lim_{x\to\infty}\frac{1}{\sqrt{\frac{x^2-2}{x^2}}} \;=\;\lim_{x\to\infty}\frac{1}{\sqrt{1-\frac{2}{x^2}}} \;=\;\frac{1}{\sqrt{1-0}} \;=\;1\)