[Limits] The given solution and my solution produce the same answer but how? Is mine correct?

[Integrate]

The solution to the limit gets to the answer but I am lost on the algebra.





How does the limit get transformed into [math]\lim_{x\to 0^+} \frac{\sqrt[]{x+1}-1}{\sqrt[]{x}\sqrt{x+1}}[/math]?

What are the missing steps?


Also I am able to achieve the correct answer but through different means. Is it by accident?


My work is as follows




Thank you

 

[Dr.Peterson]

How does the limit get transformed into [math]\lim_{x\to 0^+} \frac{\sqrt[]{x+1}-1}{\sqrt[]{x}\sqrt{x+1}}[/math]?

What are the missing steps?

...
Also I am able to achieve the correct answer but through different means. Is it by accident?

They did the same thing as you, except they recognized that the LCD is not [imath]\sqrt{x}\sqrt{x^2+x}[/imath], but [imath]\sqrt{x}\sqrt{x+1}[/imath], because [imath]\sqrt{x^2+x} = \sqrt{x}\sqrt{x+1}[/imath].

Or else they factored out [imath]\frac{1}{\sqrt{x}}[/imath] before combining fractions.

 

[BigBeachBanana]

[math]\frac{\sqrt{x^2+x}-\sqrt{x}}{\sqrt{x}\sqrt{x^2+x}}=\frac{\cancel{\sqrt{x}}(\sqrt{x+1}-1)}{\cancel{\sqrt{x}}\sqrt{x}\sqrt{x+1}}[/math]

 

[Integrate]

[math]\frac{\sqrt{x^2+x}-\sqrt{x}}{\sqrt{x}\sqrt{x^2+x}}=\frac{\cancel{\sqrt{x}}(\sqrt{x+1}-1)}{\cancel{\sqrt{x}}\sqrt{x}\sqrt{x+1}}[/math]

This definitely clicks. Now I feel dumb.

 

[Integrate]

They did the same thing as you, except they recognized that the LCD is not [imath]\sqrt{x}\sqrt{x^2+x}[/imath], but [imath]\sqrt{x}\sqrt{x+1}[/imath], because [imath]\sqrt{x^2+x} = \sqrt{x}\sqrt{x+1}[/imath].

Or else they factored out [imath]\frac{1}{\sqrt{x}}[/imath] before combining fractions.

ahhh, they pulled out the sqrt(x) sooner than I did. Got it.

Obvs in retrospect.

 

[Steven G]

From line 2 to 3 you multiplied out the denominator. I would never do that, just in case you can cancel out from top to bottom. If you distribute (ie un factor) you'll never see that things can cancel.

 

[Dr.Peterson]

From line 2 to 3 you multiplied out the denominator. I would never do that, just in case you can cancel out from top to bottom. If you distribute (ie un factor) you'll never see that things can cancel.

Agreed.

I sometimes point out that factoring is hard, distributing is easy. So you should leave things in the factored form (the higher energy state, you might say) until there's a reason to change it. Once you've rolled the stone uphill, don't push it over the edge until you've done all you could with it on the top -- and not even then!

Numerators often need to be un-factored (e.g. to add fractions); denominators almost never do. So it's a good habit to leave denominators factored.