limits (1 Viewer)

[hello-there]

How do you

Lim sin3x
x->0 5x

 

[SeCKSiiMiNh]

How do you

Lim sin3x
x->0 5x

Lim sin3x x 1/5
x->0 x

=>
Lim sin3x x 3/5
x->0 3x

= 3/5

 

[pman]

Lim sin3x x 1/5
x->0 x

=>
Lim sin3x x 3/5
x->0 3x

= 3/5

but something i noticed with sin that may save time, just cros out the sin and the x, your left with 2 numbers, thats the final solution

 

[SeCKSiiMiNh]

but something i noticed with sin that may save time, just cros out the sin and the x, your left with 2 numbers, thats the final solution

oh lol

 

[addikaye03]

but something i noticed with sin that may save time, just cros out the sin and the x, your left with 2 numbers, thats the final solution

Wait what? Explain what your saying.

lim (x-->0) (sinx)/x = 1. So you try and get it into a fraction of this so that you can use this property.

This can be proved by L' Hopitals Rule:

lim (x-->0) (sinx)/x Since function is currently indeterminable (0/0) & noting x=/=0

lim(x-->0) (cosx)/1. Now since lim(x-->0) f(x)= f(0) [Limit Law, when continous]

Therefore cos(0)/1= 1 As required

 

[hello-there]

Lim sin3x x 1/5
x->0 x

=>
Lim sin3x x 3/5
x->0 3x

= 3/5

Wait hang on if you limit x>>>>>0 sin(3x)/3x how would you get one.

 

[SeCKSiiMiNh]

Wait hang on if you limit x>>>>>0 sin(3x)/3x how would you get one.

isnt that the rule?

 

[pman]

isnt that the rule?

my mehtod gives you 3/3=1

 

[pman]

Wait what? Explain what your saying.

lim (x-->0) (sinx)/x = 1. So you try and get it into a fraction of this so that you can use this property.

This can be proved by L' Hopitals Rule:

lim (x-->0) (sinx)/x Since function is currently indeterminable (0/0) & noting x=/=0

lim(x-->0) (cosx)/1. Now since lim(x-->0) f(x)= f(0) [Limit Law, when continous]

Therefore cos(0)/1= 1 As required

the co-efficient of the x's is 1, you get 1/1=1

though as i say, doesn't work for cos but its useful to double check your answer

 

[shaon0]

Lim sin3x x 1/5
x->0 x

=>
Lim sin3x x 3/5
x->0 3x

= 3/5

Answer is 3/5!!

 

[SeCKSiiMiNh]

um...

lol

 

[shaon0]

um...

lol

Yeah, wats everyone talking about? You're correct.

 

[hello-there]

isnt that the rule?

is the rule called L hospitals rule?

anyways your right

 

[SeCKSiiMiNh]

lol i dont know about hospitals rule.

doesnt it just say in the textbook?

 

[pman]

lol i dont know about hospitals rule.

doesnt it just say in the textbook?

l'hopitals rule is uni, i do it next week

 

[hello-there]

also i got another question by its integration how do you integrate

 

[SeCKSiiMiNh]

ive never seen a question like that before in my entire life o.o

 

[hello-there]

ive never seen a question like that before in my entire life o.o

its in the cambridge textbook and its one of the first questions in the development part
+1 rep to whoeva gets this

 

[SeCKSiiMiNh]

is it exp or "e"

and ive never used cambridge

 

[shaon0]

is it exp or "e"

and ive never used cambridge

I=S e^(1/x)/x^2 dx
=-e^(1/x)+C