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- Thread starter zell_D
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- #2

LeonhardEuler

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Do you know ll'Hospital's rule?

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

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LeonhardEuler

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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.zell_D said:

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LeonhardEuler

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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?

- #9

LeonhardEuler

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No, the top will look like:

[tex]3\cdot2(2\sin{x}\cos{x}(\cos{x}^2-\sin{x}^2))[/tex]

[tex]3\cdot2(2\sin{x}\cos{x}(\cos{x}^2-\sin{x}^2))[/tex]

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hmmm what can i cancel out?

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- #11

LeonhardEuler

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

- #13

LeonhardEuler

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

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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?

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?

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- #15

lurflurf

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The suggestions so far would work, but are not the best way.zell_D said:

[tex]\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}[/tex]

Both of those limits are equal to the know limit for sin(x)/x

[tex]L_1=\lim_{x\rightarrow 0}\frac{sin(4x)}{4x}[/tex]

[tex]L_2=\lim_{x\rightarrow 0}\frac{sin(3x)}{3x}[/tex]

[tex]\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[/tex]

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- #16

LeonhardEuler

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- #17

LeonhardEuler

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Pretty clever.lurflurf said:The suggestions so far would work, but are not the best way.

[tex]\lim_{x\rightarrow 0}\frac{3\sin(4x)}{\sin(3x)}=\lim_{x\rightarrow 0}4\frac{\sin(4x)}{4x} \ \frac{3x}{\sin(3x)}[/tex]

Both of those limits are equal to the know limit for sin(x)/x

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care to explain? im sorta slow on math =/ this method is in my book i think but i do not get it lol

- #19

LeonhardEuler

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[tex]\lim_{x\rightarrow 0}\frac{\sin{x}}{x}=1[/tex]

and that

[tex]\lim_{x\rightarrow 0}\frac{x}{\sin{x}}=1[/tex]

Now suppose, for instance 4x=t:

[tex]\lim_{t\rightarrow 0}\frac{\sin{t}}{t}=1[/tex]

[tex]=\lim_{4x\rightarrow 0}\frac{\sin{4x}}{4x}=1[/tex]

So, in this problem what lurflurf did was to multiply the equation by [itex]\frac{x}{x}[/itex], which is always 1 as long as x is not 0, so it does not change the value. Then it becomes:

[tex]\lim_{x\rightarrow 0}\frac{3x\sin{4x}}{x\sin{3x}}[/tex]

[tex]=\lim_{x\rightarrow 0}\frac{3\sin{4x}}{x}\frac{x}{\sin{3x}}[/tex]

[tex]=\lim_{x\rightarrow 0}\frac{\sin{4x}}{x}\frac{3x}{\sin{3x}}[/tex]

Now he multiplies by 1 as 4/4:

[tex]=\lim_{x\rightarrow 0}4\frac{\sin{4x}}{4x}\frac{3x}{\sin{3x}}[/tex]

The limit of the product is the product of the limits, so:

[tex]=4\lim_{x\rightarrow 0}\frac{\sin{4x}}{4x}\lim_{x\rightarrow 0}\frac{3x}{\sin{3x}}[/tex]

And the answer is apparant.

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and lasty, i am suppose to state whether this statment is true or false with this graph, obviously i cant 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?

- #21

lurflurf

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That would mean for any point on the graph (c,f(c)) -1<c<1zell_D said:

and lasty, i am suppose to state whether this statment is true or false with this graph, obviously i cant 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?

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.

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