1. ## Limits question

Limit[x^0.5 ((x + 4)^0.5 - x^0.5), x -> ∞]

1. does not exist
2. exists and is equal to 0
3. exists and is equal to 1/2
4. exists and is equal to 2

(The answer is given as D. But how to proove it? )

2. you sure the answer isn't -2? or the limit goes to 0?

3. yes. the answer is 2. (and NOT -2)

4. why don't you try to graph it with a software? I am pretty sure the limit approches to -2, if there is no mistake in the paranthesis,

5. yes. graphing software show that the limit appproaches 2.

Plot[x^0.5 (-x^0.5 + (4 + x)^0.5`), {x, -10, 10}]

6. oki, here is my solution.

I multiplied and divided the term with the conjugate of the second part of the equation. then used L'hospital. Because it is inf/inf.

7. With simple manipulations you obtain...

$\displaystyle \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x})= \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x}) \cdot \frac{\sqrt{x+4}+\sqrt{x}}{\sqrt{x+4}+\sqrt{x}} =$

$\displaystyle = \frac{4\cdot \sqrt{x}}{\sqrt{x+4}+\sqrt{x}}= \frac{4}{1+\sqrt{\frac{x+4}{x}}}$

But...

$\displaystyle \lim_{x \rightarrow \infty} \sqrt{\frac{x+4}{x}}=1$

... so that...

$\displaystyle \lim_{x \rightarrow \infty} \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x})= 2$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. impressed with the use of symbols.

9. Originally Posted by sench
impressed with the use of symbols.
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