How can I find the limit 0/0 when using direct substitution?

[Cal124]

I am really struggling with limits at the moment. Any help would be great! Thanks to anyone in advance if they take the time to read the rest of this.
Basically i am struggling with finding the limit when using direct substitution provides 0/0
I (think) am fine with limits that involve quadratic equations, you factorise which should allow some canceling out, and direct substitution should provide an answer.

Just to check
Lim t->(-1) [(1/(t+1)) - (1/(t^2+3t+2)]
factorizing the t^2+3t+2 i can cancel out to 1/(t+2) which with direct substitution the limit = 1/1 =1

but my problem is when i encounter limits such as
lim h->0 (sqrt(2+2h)-sqrt(2)) / h

I did try multiplying it by the conjugate (sqrt(2+2h)-sqrt(2)) / (sqrt(2+2h)+sqrt(2)) but this gave me 2/sqrt(2) and putting the original into worlfram alpha the answer is apparently 1/sqrt(2)

I'm just not sure on the procedure when dealing with this king of limit. any help would be greatly appreciated!
Thanks

 

[micromass]

Tell us how you obtained $\frac{2}{\sqrt{2}}$. Your method is good, but there is likely a computation error.

 

[Cal124]

Tell us how you obtained $\frac{2}{\sqrt{2}}$. Your method is good, but there is likely a computation error.

Hopefully this will be okay to follow apologies if not, not sure how to write equations on here yet.
>>
[sqrt(2+2h) - sqrt(2)]/h . [sqrt(2+2h)+sqrt(2)]/sqrt(2+2h)+sqrt(2)
>>
2+2h+[sqrt(2)sqrt(2+2h)]-[sqrt(2)sqrt(2+2h)]-2
h.sqrt(2+2h)+sqrt(2)
>>
cancels to
2h
h.[sqrt(2+2h)+sqrt(2)]
>>
giving, as h->0 & canceling the h's
2
sqrt(2+2.0)+sqrt(2)
Ive realized my error, thanks a lot i will continue just in case
>>
2/sqrt(2)+sqrt(2) = sqrt(2)/2

just for clarification, is it a rule of thumb in this sort of limit to multiply by the conjugate?

 

[micromass]

Nice work!

just for clarification, is it a rule of thumb in this sort of limit to multiply by the conjugate?

Yes, it is a trick that works in a lot of cases.

 

[Cal124]

Nice work!
Yes, it is a trick that works in a lot of cases.

Thanks! Much appreciated.

 

[WWGD]

Have you tried setting $2h=u$ to make the expression look more like a derivative?

 

[SammyS]

Have you tried setting $2h=u$ to make the expression look more like a derivative?

It looks like it could be a derivative just as it is.

ƒ'(1) , where ƒ(x) = √(2x) .