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Solving trigonometric limits

  • Thread starter mtayab1994
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1. Homework Statement

solve the following limit: [tex]\lim_{x\rightarrow0}\frac{x+sinx}{2x-sin3x}[/tex]


3. The Attempt at a Solution

I know the principle of solving all sorts of limits but not trig limits, and since we just started trig limits it's still not clear to me, but i think to solve it we have to separate it then solve but i don't know. Any help on how to start please?
 

Dick

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Do you know l'Hopital's rule?
 
584
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No haven't learned it can you fill me on it please?
 

SammyS

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1. Homework Statement

solve the following limit: [tex]\lim_{x\rightarrow0}\frac{x+sinx}{2x-sin3x}[/tex]

3. The Attempt at a Solution

I know the principle of solving all sorts of limits but not trig limits, and since we just started trig limits it's still not clear to me, but i think to solve it we have to separate it then solve but i don't know. Any help on how to start please?
Are you familiar with Taylor series ?

Added in Edit:

Better yet:
multiply by (1/x) over (1/x) .​
 

Dick

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No haven't learned it can you fill me on it please?
You may not have covered it yet so you may be expected to do it a different way. Divide the numerator and denominator by x. Now you probably do know the limit sin(x)/x. Try to use that to find the limits of the terms.
 
584
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Are you familiar with Taylor series ?
No sorry i'm not familiar with taylor series, but i know that for l'hospital's rule we have to take the derivative of the numerator and the derivative of the denominator; but i have never learned derivatives so i can't do that. Is it possible to solve it some other way?
 
584
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Ok the derivative of x+sinx is cosx+1 and how about the bottom?
 
584
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ok i counted and found the limit to be -2 is that correct?
 

Dick

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ok i counted and found the limit to be -2 is that correct?
Yes, if you used l'Hopital then you might check whether that is allowed. Otherwise you should try doing it using that the limit of sin(x)/x is 1.
 
584
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Yes, if you used l'Hopital then you might check whether that is allowed. Otherwise you should try doing it using that the limit of sin(x)/x is 1.
Yea thank you very much I get the trick. It's the same for almost all of them.
 

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