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Finding the limit of a difference quotient defined by f(x) = 4/(7+3x)

Hey guys, My first post here, but basically I have this problem, and I can get the answer out, but I cheated to get the answer, my working out isn't quite correct.

I just can't work it out, I've gotten this far, but maybe I'm missing something so simple.

Attachment 28061

Anyhelp would be appreciated. I know what the answer is supposed to be, but I have a spare h left over and to get the 2*sqrt(3+5x) can't have the h there.

Re: Finding the limit of a difference quotient defined by f(x) = 4/(7+3x)

After line 6 you should see that ( 3 + 5x + 5h ) - ( 3 + 5x ) = 3 + 5x + 5h - 3 - 5x = 5h so line 7 should be

So the h that destroys numerator and denominator can cancel and the h under the radical --> 0.

:)

Re: Finding the limit of a difference quotient defined by f(x) = 4/(7+3x)

I've multiplied h by both parts, so I had an extra h before sqrt(3+5x)

Was h supposed to be multiplied by the whole part on the right, rather than both parts individually?

Thats why I had a h left over after cancelling out.

Re: Finding the limit of a difference quotient defined by f(x) = 4/(7+3x)

Quote:

Originally Posted by

**limitofx** I've multiplied h by both parts, so I had an extra h before sqrt(3+5x)

Was h supposed to be multiplied by the whole part on the right, rather than both parts individually?

Thats why I had a h left over after cancelling out.

Yes , but you didn't have to multiply it out in your line 5, notice the parenthesis in the denominator of my post?

In general, do not distribute the h in the denominator... it will cancel if you do the algebra correctly.

:)

Re: Finding the limit of a difference quotient defined by f(x) = 4/(7+3x)

thanks, I guess I just wanted to put in all the steps. I can see that all we're doing is removing the roots and multiplying h by the conjugate of the numerator. I've worked out the mistake though, and it makes sense.