# Limits with fractions

• Apr 12th 2010, 09:34 AM
CSG18
Limits with fractions
Hi,

I do not understand how the two limits are equal to each other. Can someone include more steps between them please?

I do understand that the two fractions need to be subtracted from each other, therefore the denominator must be multiplied to each part of the fraction. But why do one of the 8's disappear?

Thanks
• Apr 12th 2010, 09:51 AM
harish21
Quote:

Originally Posted by CSG18
Hi,

I do not understand how the two limits are equal to each other. Can someone include more steps between them please?

I do understand that the two fractions need to be subtracted from each other, therefore the denominator must be multiplied to each part of the fraction. But why do one of the 8's disappear?

Thanks

You don't understand that???!!

You should have done enough algebra to know that

$\frac{a}{b} - \frac{c}{d} = \frac{ad-bc}{bd}$

Likewise:

$\frac{8}{(x+h)^2} - \frac{8}{x^2} = \frac{8(x^2) - 8(x+h)^{2}}{x^{2}(x+h)^{2}}$

do you understand now?
• Apr 12th 2010, 09:54 AM
CSG18
thanks, I completely understand what you wrote, but its not what the image says....

Am I missing something, or is the image wrong?
• Apr 12th 2010, 09:55 AM
CSG18

.....(Giggle)
• Apr 12th 2010, 09:58 AM
harish21
Quote:

Originally Posted by CSG18
thanks, I completely understand what you wrote, but its not what the image says....

Am I missing something, or is the image wrong?

No, the image is not wrong!

another example: $2x-2y = 2(x-y)$

Likewise for this fraction,

$\frac{8}{(x+h)^2} - \frac{8}{x^2} = \frac{8(x^2) - 8(x+h)^{2}}{x^{2}(x+h)^{2}}$

look at the numerator:

$8(x^2) - 8(x+h)^{2} = 8({x^{2}}-(x+h)^{2}$

Clear?