Limits to infinity

**PhizKid** · October 27th 2012, 02:55 AM

$\lim_{x \to \infty} \frac{2x}{\sqrt{x+2} + \sqrt{x}}\\\\\\ \lim_{x \to \infty} \frac{\frac{2x}{x}}{\sqrt{\frac{x}{x}+\frac{2}{x}} + \sqrt{\frac{x}{x}}}\\\\\\ \lim_{x \to \infty} \frac{2}{\sqrt{1 + \frac{2}{x}} + \sqrt{1}}\\\\\\ \lim_{x \to \infty} \frac{2}{\sqrt{1} + \sqrt{1}}\\\\\\ \lim_{x \to \infty} \frac{2}{1 + 1} = \frac{2}{2} = 1\\\\\\$

I don't understand where I did my work wrong. I divided all the terms by a common multiple to simplify them, so I don't see anywhere I could have done it incorrectly.

**MarkFL** · October 27th 2012, 03:00 AM

In your first step, when you bring 1/x under the radicals, it needs to be 1/x^2.

**johnsomeone** · October 27th 2012, 03:24 AM

Originally Posted by PhizKid

$\lim_{x \to \infty} \frac{2x}{\sqrt{x+2} + \sqrt{x}}\\\\\\ \lim_{x \to \infty} \frac{\frac{2x}{x}}{\sqrt{\frac{x}{x}+\frac{2}{x}} + \sqrt{\frac{x}{x}}}$

That's where you made a mistake.

Doing it your way would be doing it like this:

$\lim_{x \to \infty} \frac{2x}{\sqrt{x+2} + \sqrt{x}}$

$= \lim_{x \to \infty} \frac{\frac{2x}{x}}{\left( \frac{1}{x} \right) \left(\sqrt{x + 2} + \sqrt{x}\right)}$

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

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

$= \lim_{x \to \infty} \frac{2}{\sqrt{\frac{1}{x^2}(x + 2)} + \sqrt{\frac{1}{x^2}(x)}}$

$= \lim_{x \to \infty} \frac{2}{\sqrt{\frac{x+2}{x^2}} + \sqrt{ \frac{x}{x^2} } }$

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

$= \lim_{x \to \infty} \frac{2}{\sqrt{ \frac{1}{x} + \frac{2}{x^2} } + \sqrt{ \frac{1}{x} } }.$

So it's like 2/0, which goes to positive infinity.

Another way (a better way I think) is to divide numerator and denominator by $\sqrt{x}$ to kill the "to infinity" happening in the denominator.

$\lim_{x \to \infty} \frac{2x}{\sqrt{x+2} + \sqrt{x}}$

$=\lim_{x \to \infty} \left( \frac{2x}{\sqrt{x+2} + \sqrt{x}} \right) \left( \frac{ \frac{1}{\sqrt{x}} }{ \frac{1}{\sqrt{x}} } \right)$

$=\lim_{x \to \infty} \left( \frac{2x \left(\frac{1}{\sqrt{x}} \right) }{ (\sqrt{x+2} + \sqrt{x})\left(\frac{1}{\sqrt{x}} \right) } \right)$

$=\lim_{x \to \infty} \frac{ \frac{2x}{\sqrt{x}} }{ \frac{\sqrt{x+2}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{x}} }$

$=\lim_{x \to \infty} \frac{ 2\sqrt{x} }{ \sqrt{ \frac{x+2}{x} } + 1 }$

$=\lim_{x \to \infty} \frac{ 2\sqrt{x} }{ \sqrt{ 1 + \frac{2}{x} } + 1 }.$

$\text{So the numerator goes to infinity, the denominator goes to 2, so the limit of the quotient is positive infinity.}$

**PhizKid** · October 27th 2012, 04:08 AM

Oh, I'm still a little confused. Our professor just told us to divide out each term by a common factor or factor some common factor out of each term, so that's just what I did. I didn't think it mattered what kind of term it was but apparently it's something very different.

So you multiply every term by (1/x) every time?

In your first method, how did your 1/x become sqrt(1/x^2)?

**Plato** · October 27th 2012, 05:15 AM

Originally Posted by PhizKid

Oh,
So you multiply every term by (1/x) every time?
In your first method, how did your 1/x become sqrt(1/x^2)?

If $x>0$ then $\frac{1}{x}=\frac{1}{\sqrt{x^2}}$

**Deveno** · October 27th 2012, 06:10 AM

another approach:

we may assume x > 2, since we are taking the limit at "positive infinity".

thus:

$\sqrt{x} < \sqrt{x+2}$ , and $2x = x + x > x + 2$

so:

$\sqrt{x+2} + \sqrt{x} < 2\sqrt{x+2}$ and:

$\frac{2x}{\sqrt{2+x}+\sqrt{x}} > \frac{2x}{2\sqrt{x+2}} > \frac{x+2}{2\sqrt{x+2}} = \frac{\sqrt{x+2}\sqrt{x+2}}{2\sqrt{x+2}} = \frac{\sqrt{x+2}}{2}$

since:

$\lim_{x \to \infty} \frac{\sqrt{x+2}}{2} = \infty$

$\lim_{x \to \infty} \frac{2x}{\sqrt{2+x}+\sqrt{x}} = \infty$ (it's always (for x > 2) bigger than half the square root of x+2, right?)

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