Limit with a radical in the denominator

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jpanknin

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My Calculus textbook (Stewart, Early Transcendentals, 8e) provides the following example:

[math]\frac{n}{\sqrt{10 + n}}[/math]
And then proceeds to the following expression:

[math]\frac{1}{\sqrt{\dfrac{10}{n^2} + \dfrac{1}{n}}}[/math]
I don't understand what happened to get from the first expression to the second, specifically how

[math]\frac{\sqrt{10 + n}}{n} = \sqrt{\frac{10}{n^2} + \frac{1}{n}}[/math]
Any help is appreciated.
 
If you divide both the numerator and denominator of the first fraction by [imath]n[/imath] you will get the second fraction.
 
jpanknin said:
My Calculus textbook (Stewart, Early Transcendentals, 8e) provides the following example:

[math]\frac{n}{\sqrt{10 + n}}[/math]
And then proceeds to the following expression:

[math]\frac{1}{\sqrt{\dfrac{10}{n^2} + \dfrac{1}{n}}}[/math]
I don't understand what happened to get from the first expression to the second, specifically how

[math]\frac{\sqrt{10 + n}}{n} = \sqrt{\frac{10}{n^2} + \frac{1}{n}}[/math]
Any help is appreciated.
Click to expand...
Dividing the radical by n divides the radicand by [imath]n^2[/imath].

If you don't see that, you might try thinking through it in reverse: If you were given the RHS, how would you simplify it? Many processes you'll run across in calculus are the reverse of processes you have probably seen, such as simplifying.
 
That did help @Dr.Peterson. I was able to reverse it from:

[math]\sqrt{\frac{10}{n^2} + \frac{1}{n}}...to...\frac{\sqrt{10 + n}}{n}[/math]
But when I try to go the other way, I lose the square root on the top:

[math]\frac{\sqrt{10 + n}}{n}...==>...\frac{(\sqrt{10 + n})^2}{(n)^2}...==>...\frac{10 + n}{n^2}...==>...\frac{10}{n^2} + \frac{1}{n}[/math]
What am I missing here?
 
jpanknin said:
That did help @Dr.Peterson. I was able to reverse it from:

[math]\sqrt{\frac{10}{n^2} + \frac{1}{n}}...to...\frac{\sqrt{10 + n}}{n}[/math]
But when I try to go the other way, I lose the square root on the top:

[math]\frac{\sqrt{10 + n}}{n}...==>...\frac{(\sqrt{10 + n})^2}{(n)^2}...==>...\frac{10 + n}{n^2}...==>...\frac{10}{n^2} + \frac{1}{n}[/math]
What am I missing here?
Click to expand...
You can't square an expression when you are simplifying it! That changes the value of the expression. But if you do square it, then take the square root of what you end up with ...

Here is how I would do it: [math]\frac{\sqrt{10 + n}}{n}=\frac{\sqrt{10 + n}}{\sqrt{n^2}}=\sqrt{\frac{10 + n}{n^2}}=\sqrt{\frac{10}{n^2} + \frac{1}{n}}[/math]
 
Assuming n > 0

[math]\dfrac{n}{\sqrt{10 + n}} = \dfrac{n}{\sqrt{10 + n}} * \dfrac{1/n}{1/n} = \dfrac{1}{\dfrac{\sqrt{10 + n}}{n}} =[/math]
[math]\dfrac{1}{\dfrac{\sqrt{10 + n}}{\sqrt{n^2}}} = \dfrac{1}{\sqrt{\dfrac{10 + n}{n^2}}} =\dfrac{1}{\sqrt{\dfrac{10}{n} + \dfrac{1}{n}}}.[/math]
 
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