Limits - square root

[Rectifier]

The problem
$$ \lim_{x \rightarrow \infty} \left( \sqrt{x+1} - \sqrt{x} \right) $$

The attempt
$\left( \sqrt{x+1} - \sqrt{x} \right) = \frac{\left( \sqrt{x+1} - \sqrt{x} \right)\left( \sqrt{x+1} + \sqrt{x} \right) }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{x+1 - x }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{1 }{x \left( \sqrt{\frac{1}{x}+\frac{1}{x^2}} + \sqrt{\frac{1}{x}} \right) } = \frac{1}{x} \frac{1 }{ \sqrt{\frac{1}{x}+\frac{1}{x^2}} + \sqrt{\frac{1}{x}} } \\ \rightarrow 0 \cdot \frac{1}{0 + 0}$ for $x \rightarrow \infty$ division with 0 is undefined. Can somone help me calculate this limit.

And no, I can't use l'Hopitals rule so please don't mention it in this thread.

 

[member 587159]

The problem
$$ \lim_{x \rightarrow \infty} \left( \sqrt{x+1} - \sqrt{x} \right) $$

The attempt
$\left( \sqrt{x+1} - \sqrt{x} \right) = \frac{\left( \sqrt{x+1} - \sqrt{x} \right)\left( \sqrt{x+1} + \sqrt{x} \right) }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{x+1 - x }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{1 }{x \left( \sqrt{\frac{1}{x}+\frac{1}{x^2}} + \sqrt{\frac{1}{x}} \right) } = \frac{1}{x} \frac{1 }{ \sqrt{\frac{1}{x}+\frac{1}{x^2}} + \sqrt{\frac{1}{x}} } \\ \rightarrow 0 \cdot \frac{1}{0 + 0}$ for $x \rightarrow \infty$ division with 0 is undefined. Can somone help me calculate this limit.

And no, I can't use l'Hopitals rule so please don't mention it in this thread.

$\left( \sqrt{x+1} - \sqrt{x} \right) = \frac{\left( \sqrt{x+1} - \sqrt{x} \right)\left( \sqrt{x+1} + \sqrt{x} \right) }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{1 }{\left( \sqrt{x+1} + \sqrt{x} \right) }$.

Until here, what you do leads to the answer. When you fill in infinity right now, you don't get an undetermined form, so the answer is ...?

 

[Ray Vickson]

The problem
$$ \lim_{x \rightarrow \infty} \left( \sqrt{x+1} - \sqrt{x} \right) $$

The attempt
$\left( \sqrt{x+1} - \sqrt{x} \right) = \frac{\left( \sqrt{x+1} - \sqrt{x} \right)\left( \sqrt{x+1} + \sqrt{x} \right) }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{x+1 - x }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{1 }{x \left( \sqrt{\frac{1}{x}+\frac{1}{x^2}} + \sqrt{\frac{1}{x}} \right) } = \frac{1}{x} \frac{1 }{ \sqrt{\frac{1}{x}+\frac{1}{x^2}} + \sqrt{\frac{1}{x}} } \\ \rightarrow 0 \cdot \frac{1}{0 + 0}$ for $x \rightarrow \infty$ division with 0 is undefined. Can somone help me calculate this limit.

And no, I can't use l'Hopitals rule so please don't mention it in this thread.

Note that
$$\frac{x+1-x}{\sqrt{x}+\sqrt{x+1}} = \frac{1}{\sqrt{x}+\sqrt{x+1}}$$
In the second form, what happens to the numerator when $x \to +\infty$? What happens to the denominator?

What dictator forbids you from using l'Hospital's rule?

 

[Mark44]

What dictator forbids you from using l'Hospital's rule?

I haven't worked this problem, but sometimes L'Hopital's rule doesn't get you anywhere. I've seen similar problems where two applications of L'H gets you right back to where you started.