MHB What is the Limit of a Rational Expression?

[karush]

$$\lim_{{x}\to{\infty}}\frac{\sqrt{3x^2 - 1 }}{x-1}=\sqrt{3}$$

I tried dividing everything by ${x}^{2}$ but not

 

[evinda]

$$\lim_{{x}\to{\infty}}\frac{\sqrt{3x^2 - 1 }}{x-1}=\sqrt{3}$$

I tried dividing everything by ${x}^{2}$ but not

$$\lim_{x \to +\infty} \frac{\sqrt{3x^2-1}}{x-1}= \lim_{x \to +\infty} \frac{3x^2-1}{(x-1) \sqrt{3x^2-1}}= \lim_{x \to +\infty} \frac{x^2 \left( 3- \frac{1}{x^2}\right)}{(x-1) \sqrt{x^2 \left( 3-\frac{1}{x^2}\right)}}= \lim_{x \to +\infty} \frac{x^2 \left( 3- \frac{1}{x^2}\right)}{(x-1)|x| \sqrt{ \left( 3-\frac{1}{x^2}\right)}}\\= \lim_{x \to +\infty} \frac{x^2 \left( 3- \frac{1}{x^2}\right)}{(x-1)x \sqrt{ \left( 3-\frac{1}{x^2}\right)}}= \lim_{x \to +\infty} \frac{x^2}{x^2 \left(1- \frac{1}{x} \right)} \cdot \lim_{x \to +\infty} \frac{3-\frac{1}{x^2}}{\sqrt{3-\frac{1}{x^2}}}= 1\cdot \frac{3}{\sqrt{3}}= \sqrt{3}$$

 

[karush]

Wow, thanks, that was a lot of steps.
How would you know initially to go in that direction.

 

[evinda]

Wow, thanks, that was a lot of steps.

(Smile)

How would you know initially to go in that direction.

Usually when you have a square root you multiply by it at the numerator and the denominator.

 

[karush]

Could you move this thread to the correct category?

 

[Prove It]

$$\lim_{{x}\to{\infty}}\frac{\sqrt{3x^2 - 1 }}{x-1}=\sqrt{3}$$

I tried dividing everything by ${x}^{2}$ but not

The easier way (first note that I'm leaving out absolute values because everything is positive when going to infinity anyway...)

$\displaystyle \begin{align*} \frac{\sqrt{3x^2 - 1}}{x - 1} &= \frac{\frac{1}{x}\,\sqrt{3x^2 - 1}}{\frac{1}{x} \left( x - 1 \right) } \\ &= \frac{\frac{1}{\sqrt{x^2}}\,\sqrt{3x^2 - 1}}{1 - \frac{1}{x}} \\ &= \frac{\sqrt{\frac{3x^2 - 1}{x^2}}}{1 - \frac{1}{x}} \\ &= \frac{\sqrt{3 - \frac{1}{x^2}}}{1 - \frac{1}{x}} \end{align*}$

So as $\displaystyle \begin{align*} x \to \infty , \, \frac{1}{x} \to 0 \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{x^2} \to 0 \end{align*}$, giving $\displaystyle \begin{align*} \frac{\sqrt{3 - 0}}{1 - 0} = \sqrt{3} \end{align*}$ as the limit :)

 

[karush]

I actually got to your 3rd step before posted, just didn't see the magic.

I sure learn a lot with MHB