Understanding the Definition of a Limit in Graphs

[zell_D]

lim x->0 3sin4x/sin3x i do not know how to reduce the sin4x? and do i even use the property where lim x->0 sinx/x =1?

 

[LeonhardEuler]

Do you know ll'Hospital's rule?

 

[zell_D]

never heard of it, class just started so if i learned anything i forgot

 

[LeonhardEuler]

If you have'nt done it, then don't use it. Use the formulas for the sine and cosine of double angles and sums of angles (i.e. use the sum of angles to make sin(3x)=sin(x+2x), into an expresion with only x and 2x, then use the double angle rules to make the functions of 2x into functions of x. Simmilarly, 4x=2(2x).)

 

[zell_D]

ok thanks another REAL DUMB question on my part lol haven't done math for so long, when i make sin4x into sin2(2x) and then into 2sin2xcos2x, do i use distributive property with the 3?

 

[fourier jr]

ok thanks another REAL DUMB question on my part lol haven't done math for so long, when i make sin4x into sin2(2x) and then into 2sin2xcos2x, do i use distributive property with the 3?

no, there's an identity that's probably on the inside cover of your textbook. it goes something like sin(a+b) = cos(a)sin(b) + cos(b)sin(a). i can't remember how that formula goes but it's something like that.

 

[LeonhardEuler]

Yes, Fourier has it right. sin(3x)=sin(x+2x)=sin(x)cos(2x)+sin(2x)cos(x), then use the rule for sin(2x) again, and the rule cos(2x)=cos^2(x)-sin^2(x).

 

[zell_D]

hmm can anyone check my work lol i feel dumb and still can't get the answer =/

lim x-> 0(from here on stated as lim)


lim 3sin4x/sin3x
= lim 3sin2(2x)/sin(x+2x)
= lim (3)2sin2xcos2x/sinxcos2x+cosxsin2x

what do i do after this?
do i need to use the other double angle property so the top looks like sin2xcos2x+cos2xsin2x?

 

[LeonhardEuler]

No, the top will look like:
3\cdot2(2\sin{x}\cos{x}(\cos{x}^2-\sin{x}^2))

 

[zell_D]

hmmm what can i cancel out?

 

[LeonhardEuler]

You just need to simplify it a little more with the double angle formulas in the denomenator to get a solution.

 

[zell_D]

does it matter which form of cos' double angle formula i use?

 

[LeonhardEuler]

No, it will take about the same amount of work whichever way.

 

[zell_D]

lol... i ran outta room and still not finishing =/

do i use distributive property for the cosx^2-sinx^2?

ok i finished the answer is 4 is that correct? also i am still confused on one part, any confirmation would b nice


From:

lim 12sinxcosx(cosx^2-sinx^2)/3sinxcos^2(x)-sin^3(x)<-- can anyone check math on this part?
=lim 12sinxcos^3(x)-12sin^3(x)cosx/3sinxcos^2(x)-sin^3(x)
now is my problem, assuming i did i tall correctly, i know i factor sinx from the bottom, but how bout the 3?

 

[lurflurf]

lim x->0 3sin4x/sin3x i do not know how to reduce the sin4x? and do i even use the property where lim x->0 sinx/x =1?

The suggestions so far would work, but are not the best way.
\lim_{x\rightarrow 0}\frac{3\sin(4x)}{\sin(3x)}=\lim_{x\rightarrow 0}4\frac{\sin(4x)}{4x} \ \frac{3x}{\sin(3x)}=4\frac{L_1}{L_2}
Both of those limits are equal to the know limit for sin(x)/x
L_1=\lim_{x\rightarrow 0}\frac{sin(4x)}{4x}
L_2=\lim_{x\rightarrow 0}\frac{sin(3x)}{3x}
\lim_{x\rightarrow 0}\frac{sin(4x)}{4x}=\lim_{x\rightarrow 0}\frac{sin(3x)}{3x}=\lim_{x\rightarrow 0}\frac{sin(x)}{x}=1

 

[LeonhardEuler]

Not much further to go after you make those substitutions. Then you will be able to factor out a sin(x) that will leave some of the terms with only cos(x). Then you can take the limit, all the terms with sin(x) go to zero, all the terms with only cos(x) go to 1, and you have your answer.

 

[LeonhardEuler]

The suggestions so far would work, but are not the best way.
\lim_{x\rightarrow 0}\frac{3\sin(4x)}{\sin(3x)}=\lim_{x\rightarrow 0}4\frac{\sin(4x)}{4x} \ \frac{3x}{\sin(3x)}
Both of those limits are equal to the know limit for sin(x)/x

Pretty clever.

 

[zell_D]

care to explain? I am sort of slow on math =/ this method is in my book i think but i do not get it lol

 

[LeonhardEuler]

Ok, you know that
\lim_{x\rightarrow 0}\frac{\sin{x}}{x}=1
and that
\lim_{x\rightarrow 0}\frac{x}{\sin{x}}=1
Now suppose, for instance 4x=t:
\lim_{t\rightarrow 0}\frac{\sin{t}}{t}=1
=\lim_{4x\rightarrow 0}\frac{\sin{4x}}{4x}=1
So, in this problem what lurflurf did was to multiply the equation by \frac{x}{x}, which is always 1 as long as x is not 0, so it does not change the value. Then it becomes:
\lim_{x\rightarrow 0}\frac{3x\sin{4x}}{x\sin{3x}}
=\lim_{x\rightarrow 0}\frac{3\sin{4x}}{x}\frac{x}{\sin{3x}}
=\lim_{x\rightarrow 0}\frac{\sin{4x}}{x}\frac{3x}{\sin{3x}}
Now he multiplies by 1 as 4/4:
=\lim_{x\rightarrow 0}4\frac{\sin{4x}}{4x}\frac{3x}{\sin{3x}}
The limit of the product is the product of the limits, so:
=4\lim_{x\rightarrow 0}\frac{\sin{4x}}{4x}\lim_{x\rightarrow 0}\frac{3x}{\sin{3x}}
And the answer is apparant.

 

[zell_D]

o wow that was much ezier thanks a lot guys, sorry for all the bothering =/

and lasty, i am suppose to state whether this statement is true or false with this graph, obviously i can't draw the graph on here but would one of you tell me what does:

lim x-> c f(x) exists at every c in (-1, 1) mean?

 

[lurflurf]

o wow that was much ezier thanks a lot guys, sorry for all the bothering =/

and lasty, i am suppose to state whether this statement is true or false with this graph, obviously i can't draw the graph on here but would one of you tell me what does:

lim x-> c f(x) exists at every c in (-1, 1) mean?

That would mean for any point on the graph (c,f(c)) -1<c<1
you could draw horizontal lines
y=f(c)+h
y=f(c)-h
with h>0 (but often we imagine it is small)
it would then be possible to draw vertical lines
x=a
x=b
with a<c<b
so that no point on the graph with x between a and b (a<x<b)
is outside the rectangle defined by
a<x<b
f(c)-h<y<f(c)+h

in practice look for a break in the graph
so that it looks like a small rectangle drawn around a point would have some graph above or below it.