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Determine limit as x approaches 0?

Im quite stuck on one question in my most recent assignment, where it asks me to :

Determine :

CodeCogsEqn.gif

I'm confused to as to what the question is asking me/ where i should start(never really understood limits)
? I dont want anyone to walk me through it, but if someone could point me in the right direction it would be appreciated
Reply 1
I would use the taylor series for cos x here.
Reply 2
Original post by DFranklin
I would use the taylor series for cos x here.


Alright so my problem is i dont know where the rest of the equation fits in? what would i do with the x^2/2 and the x^4 ? I'm assuming the x^4 means that i expand the series to the 4th term?
Original post by santeria133
Alright so my problem is i dont know where the rest of the equation fits in? what would i do with the x^2/2 and the x^4 ? I'm assuming the x^4 means that i expand the series to the 4th term?


and then divide the result by the x^4 in the denominator
Reply 4
Keep applying L'Hopitals rule.
Reply 5
Original post by lubus
Keep applying L'Hopitals rule.


The Taylor series approach is much simpler :smile:
Reply 6
Original post by santeria133
Im quite stuck on one question in my most recent assignment, where it asks me to :

Determine :

CodeCogsEqn.gif

I'm confused to as to what the question is asking me/ where i should start(never really understood limits)
? I dont want anyone to walk me through it, but if someone could point me in the right direction it would be appreciated


Bound the function.
(edited 10 years ago)
Reply 7
Original post by santeria133
Im quite stuck on one question in my most recent assignment, where it asks me to :

Determine :

CodeCogsEqn.gif

I'm confused to as to what the question is asking me/ where i should start(never really understood limits)
? I dont want anyone to walk me through it, but if someone could point me in the right direction it would be appreciated


While the limit of both the numerator and denominator is 0, you can use the
L'Hospital rule (3 times),then use that

limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x}=1

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