anonymous
  • anonymous
Limit help please!! What is the limit of ((1/(x+ deltax)-(1/x))/deltax as deltax approaches 0 from the left?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
blockcolder
  • blockcolder
Try simplifying the numerator.
anonymous
  • anonymous
I'm not sure how to.. I know when you subtract/add fractions you need a common denominator, but if you do that for the big numerator, do you also have to do it for the big denominator?
blockcolder
  • blockcolder
Not really. When you work on the numerator, ignore the denominator for the meantime. After simplifying the numerator, put the denominator back.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
When I simplified the numerator and put the denominator back, I got delta x divided by (x + delta x) divided by delta x. Is that right so far?
blockcolder
  • blockcolder
\[\frac{\frac{\Delta x}{x+\Delta x}}{\Delta x}\ ?\]
anonymous
  • anonymous
yeah
blockcolder
  • blockcolder
That's not quite right. \[\frac{1}{x+\Delta x}-\frac{1}{x}=\frac{x-(x+\Delta x)}{x(x+\Delta x)}\]
anonymous
  • anonymous
when i found the common denominator i multiplied the first fraction by 1 and the second fraction by (1+ delta x)
blockcolder
  • blockcolder
But \[\frac{1}{x}\cdot\frac{1}{1+\Delta x}=\frac{1}{x(1+\Delta x)}=\frac{1}{x+x\Delta x}\] doesn't have the same denominator as \(\frac{1}{x+\Delta x}\).
anonymous
  • anonymous
Oh. I see what I did wrong, but what should I multiply it by?
blockcolder
  • blockcolder
You can multiply it by \(1/(x+\Delta x)\). Then you multiply the first factor by \(1/x\).
anonymous
  • anonymous
and when you simplify it the answer is zero?
anonymous
  • anonymous
Is it \[\huge \lim_{h \rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}\]
blockcolder
  • blockcolder
No. It will be \[\frac{1}{x+\Delta x}-\frac{1}{x}=\frac{x-(x+\Delta x)}{x(x+\Delta x)}=\frac{-\Delta x}{x(x+\Delta x)}\]
anonymous
  • anonymous
Can you do it step by step? I'm not getting any of this
blockcolder
  • blockcolder
\[\begin{align} \frac{1}{x+\Delta x}-\frac{1}{x}&=\frac{x}{x(x+\Delta x)}-\frac{x+\Delta x}{x(x+\Delta x)}\\ &=\frac{x-(x+\Delta x)}{x(x+\Delta x)}\\ &=\frac{x-x-\Delta x}{x(x+\Delta x)}\\ &=\frac{-\Delta x}{x(x+\Delta x)}\\ \end{align}\]
anonymous
  • anonymous
I think I'll just get my teacher to explain it - I'm just getting more confused :/

Looking for something else?

Not the answer you are looking for? Search for more explanations.