Limit Problem: Solving \lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}

[Ssnow]

ok, the first thing to do in order to determine $\lim_ {x\rightarrow 0} \frac{\sqrt{x}(x-7)}{\sqrt{x}-\sqrt{7}}$ is to substitute the value $0$ in the espression $\frac{\sqrt{x}(x-7)}{\sqrt{x}-\sqrt{7}}$ and see what happen ...

 

[gede]

$\lim_{x \to 0} \frac{\sqrt{x}(x - 7)}{\sqrt{x} - \sqrt{7}}$
$=\frac{\sqrt{0}(0 - 7)}{\sqrt{0} - \sqrt{7}}$
$=\frac{0 (- 7)}{- \sqrt{7}}$
$= 0$

Then what?

 

[Ssnow]

Then the limit is $0$

 

[Mark44]

All it is correct but not necessary, I suggest you to put $x=0$ in the original limit, there are indefinite forms as $\frac{0}{0},\frac{\infty}{\infty}, 0\cdot \infty, \infty-\infty$ or not ?

In English these are called indeterminate forms.

 

[Mark44]

How to solve this limit?

\lim_{x \to 0} \frac{\sqrt{x} (x - 7)}{\sqrt{x} - \sqrt{7}}

This is what I get:

\lim_{x \to 0} \frac{\sqrt{x} (x - 7)}{\sqrt{x} - \sqrt{7}} \frac{\sqrt{x} + \sqrt{7}}{\sqrt{x} + \sqrt{7}}
= \lim_{x \to 0} \frac{\sqrt{x} (x - 7) (\sqrt{x} + \sqrt{7})}{(x - 7)}
= \lim_{x \to 0} \sqrt{x}(\sqrt{x} + \sqrt{7})

What is the next solution?

The limit on the first line above can be evaluated merely by substituting x = 0. You don't need to any of the work you show on the following lines.

 

[Ssnow]

In English these are called indeterminate forms.

Yes sorry, an error in the translation ...