1. ## Limit thread number 2?

Can anyone help me with these two limit not using l hospital rule/.

1. As x approaches 1

(1/x)-1/(x-1)

2. as h approaches zero

(2h^3-h^2+5h)/(h)

2. ## Re: Limit thread number 2?

Originally Posted by homeylova223

2. as h approaches zero

(2h^3-h^2+5h)/(h)
Factor 'h' out of the numerator. What do you get?

3. ## Re: Limit thread number 2?

I get h(2h^2-h+5)/(h)

4. ## Re: Limit thread number 2?

Cancel out h from numerator and denominator then make h=0

5. ## Re: Limit thread number 2?

I got h=5 I was stuck but you make it seem so easy.

6. ## Re: Limit thread number 2?

Originally Posted by homeylova223
I got h=5

Originally Posted by homeylova223
I was stuck but you make it seem so easy.
The best way to master math is send lots of time studying. There's no such thing as too much math.

7. ## Re: Limit thread number 2?

On my first one (1/x)-1/(x-1) as x approaches 1

Would It the final answer be zero because if I plug in one that is all I seem to get.

8. ## Re: Limit thread number 2?

Nope, that is not the answer.

Your function can also be written as -1/x(x-1)

Have you tried to graph the function?

9. ## Re: Limit thread number 2?

Originally Posted by homeylova223
I got h=5 I was stuck but you make it seem so easy.
Not h= 5, the limit, as h goes to 0, is 5.

For the first one, what you wrote, (1/x)-1/(x-1), does NOT have a limit as x goes to 1, but I think what you meant was ((1/x)- 1)/(x- 1).

To find that limit, multiply both numerator and denominator by x to get (1- x)/(x(x-1)= -(x-1)/(x(x-1)). Now the x-1 terms cancel.