Question

Limits question! Will be posted below.

Answer 1

\[\lim_{x \rightarrow 0^+} \sqrt{\frac{1}{x}+2} -\sqrt{\frac{1}{x}}\]

Answer 2

seems unlikey

Answer 3

I was thinking about it for a bit, too, but the answer given is 0.

Answer 4

oh doe i am silly

Answer 5

multiply by the conjugate will probably work

Answer 6

\[\lim_{x \rightarrow 0^+}\frac{2} {\sqrt{\frac{1}{x}+2} +\sqrt{\frac{1}{x}}}\]

Is that right when it's multiplied by its conjugate?

Answer 7

Yes, that's correct, and now you can tell the denominator gets approaches infinity as both its terms approach infinity and the summed.

Answer 8

Why is the answer given 0, though?

Answer 9

\[\lim_{x \rightarrow 0^+}(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}) \\ \lim_{x \rightarrow 0^+}(\sqrt{\frac{1+2x}{x}}-\sqrt{\frac{1}{x}}) \\ \text{ so we have } x>0 \\ \text{ then } 2x+1>0 \\ \lim_{x \rightarrow 0^+}(\frac{\sqrt{1+2x}}{\sqrt{x}}-\frac{1}{\sqrt{x}}) \\ \lim_{x \rightarrow 0^+} (\frac{\sqrt{x}\sqrt{1+2x}-\sqrt{x}}{x}) \text{ combined fractions } \\ \text{ rationalize numerator } \lim_{x \rightarrow 0^+} \frac{\sqrt{x}\sqrt{1+2x}-\sqrt{x}}{x} \cdot \frac{ \sqrt{x} \sqrt{1+2x}+\sqrt{x}}{\sqrt{x}\sqrt{1+2x}+\sqrt{x}} \\\]

Answer 10

I think you should be able to finish this you will have to do a little canceling after this rationalizing part

Answer 11

how about this...

\[\lim_{x \rightarrow 0^+} \sqrt{\frac{1}{x}+2} -\sqrt{\frac{1}{x}}=\lim_{n\to \infty}(\sqrt{n+2}-\sqrt{n})=\lim_{n\to\infty}\frac{2}{\sqrt{n+2}+\sqrt{n}}\]

\[\lim_{n\to\infty}(\sqrt{n+2}+\sqrt{n})=\infty\]

thus \[\lim_{n\to\infty}\frac{2}{\sqrt{n+2}+\sqrt{n}}=0\]

Answer 12

@Zarkon Ohhhhhh, this is great! Thanks so much.

Thank you to everyone else who contributed!

Answer 13

I feel like I have to show my full way now:
But yeah zarkon's is much shorter.
\[\lim_{x \rightarrow 0^+}(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}) \\ \lim_{x \rightarrow 0^+}(\sqrt{\frac{1+2x}{x}}-\sqrt{\frac{1}{x}}) \\ \text{ so we have } x>0 \\ \text{ then } 2x+1>0 \\ \lim_{x \rightarrow 0^+}(\frac{\sqrt{1+2x}}{\sqrt{x}}-\frac{1}{\sqrt{x}}) \\ \lim_{x \rightarrow 0^+} (\frac{\sqrt{x}\sqrt{1+2x}-\sqrt{x}}{x}) \text{ combined fractions } \\ \text{ rationalize numerator } \lim_{x \rightarrow 0^+} \frac{\sqrt{x}\sqrt{1+2x}-\sqrt{x}}{x} \cdot \frac{ \sqrt{x} \sqrt{1+2x}+\sqrt{x}}{\sqrt{x}\sqrt{1+2x}+\sqrt{x}} \\ \lim_{x \rightarrow 0^+}\frac{x(1+2x)-x}{x(\sqrt{x}\sqrt{1+2x}+\sqrt{x})} =\lim_{x \rightarrow 0^+}\frac{2x^2}{x \sqrt{x}(\sqrt{1+2x}+1)} \\ =\lim_{x \rightarrow 0^+}\frac{2 \sqrt{x}}{\sqrt{1+2x}+1}=0\]