Calculus Anyone(2)? ASAP!

[chosen5]

Can someone help me with calculus, pretty please? =)

Limits:
1: x -> 0

( (1/4) - ( 1 / (x+4)) ) / x

2: x -> 0

( (x+2)^1/2 - (2)^1/2) ) / x

3: x -> 0

2+(5/x^2)

First Derivatives:
1: x^3(x+1)^1/2

2: cos^4(5x^2)

3: x^2-3xy+y^2=42

 

[pjorstad]

I SURE HOPE I GET KARMA FOR THIS HEHEHE!




First one doesn't exist because plug in zero into x for numerator after you combine the 2 terms into a whole fraction and youll get a zero for the numberator and you get zero for denonminator since your going to zero so the limit can't exist since 0 divided by 0 doesn't. Graph it if you don't believe me



for first derivative of number 1

Your gonna need to use the product rule in combination with the chain rule for the (X+ 1) ^ 1/2 part but fortunately the derivative of the inside for of those parenthesis is just 1 so it can be ignored when using the chain rule so in a sense its just theproduct rule by itself.

Double check i did number 2 right as im on 6-OXO and rather incoherent




If anyone needs help on calculus 2 or 3 or any upper division or master degree math for that matter just post here ill try to help as much as i can IM KIDDING GUYS!

I

 

[pogue]

6oxo makes you incoherent? does it give you a buzz or something?

 

[pjorstad]

haha no i meant it kinda makes me down and moody.