# Math Help - Limit question

1. ## Limit question

I am wondering if there is a way to calculate limits or functions as x=> a certain point say x=>1 using derivatives? I remember there being a way to do it but I can't find any stuff online or in books

thanks

2. Hello,
Originally Posted by Calky
I am wondering if there is a way to calculate limits or functions as x=> a certain point say x=>1 using derivatives? I remember there being a way to do it but I can't find any stuff online or in books

thanks
l'Hôpital's rule - Wikipedia, the free encyclopedia

But it's not for any sort of limits. Sometimes (and often), you'll have to think by yourself

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Okay, there may be something else (since I'm not sure what you're asking for ^^') !

$\lim_{x \to a} \frac{f(x)-f(a)}{x-a}=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}=f ~'(a)$

3. Sometimes you might need to use the definition of the derivative:
$\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

To compute limits like: $\lim_{h \to 0} \frac{5^h - 2^h}{h}$

4. so this would be correct?

I don't know why but I always find wikipedia definitions and explanations hard to comprehend

5. The exponent is 1/2 not -1/2. It's already in the denominator.

EDIT: or is it?

EDIT2: You are still incorrect.

6. Originally Posted by Calky
so this would be correct?

I don't know why but I always find wikipedia definitions and explanations hard to comprehend
Do you know about conjugates ?

This is mostly about this identity : $(a-b)(a+b)=a^2-b^2$

So when you have a square root that's disturbing you, you can multiply by its conjugate, because it will square the square root.

For example, the conjugate of $3-\sqrt{x}$ is $3+\sqrt{x}$

So the trick (that you can use in many situations) is to multiply by $1=\frac{3+\sqrt{x}}{3+\sqrt{x}}$

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Remember that l'Hospital's rule is used in situations $\frac \infty \infty$ or $\frac 00$

7. I found that the derivative of 3-x^(1/2) = -1/2x^(-1/2) using the power rule is that incorrect? is that what you are saying

8. Originally Posted by Moo
Do you know about conjugates ?

This is mostly about this identity : $(a-b)(a+b)=a^2-b^2$

So when you have a square root that's disturbing you, you can multiply by its conjugate, because it will square the square root.

For example, the conjugate of $3-\sqrt{x}$ is $3+\sqrt{x}$

So the trick (that you can use in many situations) is to multiply by $1=\frac{3+\sqrt{x}}{3+\sqrt{x}}$

------------------------------------
Remember that l'Hospital's rule is used in situations $\frac \infty \infty$ or $\frac 00$

ok thank you

I did it that way and found that you would be dividing by zero.......

9. If you meant:
$-\frac{1}{2} x^{-1/2}$

Then you are correct.

10. Originally Posted by Chop Suey
If you meant:
$-\frac{1}{2} x^{-1/2}$

Then you are correct.

yes that is what I meant. maybe my notation or handwriting is a little off

thanks a lot

11. Originally Posted by Calky
ok thank you

I did it that way and found that you would be dividing by zero.......
No, because in the denominator, you'll get something that will simplify with the numerator

12. Originally Posted by Moo
No, because in the denominator, you'll get something that will simplify with the numerator

oh ok I see that now, I ended up doing it the other way

can you use l'hopital rule for piecewise functions?

for example with

f(x) = { 2x-6/x-3, x =/= 3
..........{ 5, x=3

finding limit of f(x) as x=> 3