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A few hopefully easy limits
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Riastrad
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#1
A few hopefully easy limits

1. Lim (sinx - cosxsinx) / x^2
x->0


2. Lim (tanx) / (x+(pie/4))
x->(pie/4)



3. Lim ((2x^3) + 1) / ((5x^3) + (2x^2) + 1)
x->inf.



Thank you all in advance!

PostPosted: Sun Sep 18, 2005 12:44 am  Back to top 
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liu
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#2
I will omit x\to 0 in limits.
\lim \frac{\sin x - \cos x \sin x}{x^2} = \lim \frac{\sin x (1-\cos x)}{x^2} = \lim \frac{\sin x}{x} \frac{2\sin^2 \frac{x}{2...

PostPosted: Sun Sep 18, 2005 1:25 am  Back to top 
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Arne
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#3
\lim_{x \rightarrow \frac{\pi}{4}} \frac{\tan{x}}{x + \frac{\pi}{4}} =  \frac{\tan{\frac{\pi}{4}}}{\frac{\pi}{4} + \frac{\pi}...

So, you just plug in \frac{\pi}{4}...
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PostPosted: Sun Sep 18, 2005 1:36 am  Back to top 
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Arne
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#4
The last one: \lim_{x \rightarrow \infty} \frac{2x^3 + 1}{5x^3 + 2x^2 + 1} = \frac{2}{5}.
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PostPosted: Sun Sep 18, 2005 1:38 am  Back to top 
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Riastrad
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#5
Arne wrote:
The last one: \lim_{x \rightarrow \infty} \frac{2x^3 + 1}{5x^3 + 2x^2 + 1} = \frac{2}{5}.


How did you cancle out the x's?

PostPosted: Sun Sep 18, 2005 2:01 pm  Back to top 
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Arne
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#6
The limit of the quotient of two polynomials (of equal degrees) is just the quotient of the coefficients of their highest degree terms Huh?
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PostPosted: Mon Sep 19, 2005 12:04 am  Back to top 
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Matnomi
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#7
Riastrad wrote:
Arne wrote:
The last one: \lim_{x \rightarrow \infty} \frac{2x^3 + 1}{5x^3 + 2x^2 + 1} = \frac{2}{5}.


How did you cancle out the x's?


basically divide numerator and denominator by x^3

PostPosted: Mon Sep 19, 2005 2:32 am  Back to top 
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