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June 28th, 2017, 05:47 AM   #1
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Hello everybody

I appreciate your help with this problem:

limit (((1+tan(x))^0.5-(1+sin(x))^0.5)/x^3,x->0)

I can solve this with L' Hôpital's Rule by differentiating the numerator and denominator 3 times and finding the limit of the resultant numerator and denominator when x->0 and dividing them: (3/2)/6=1/4!

But I'd like to see whether there is another way of solving this problem without using L' Hôpital's Rule, like by some trick or linearization... it was assigned in part of the textbook that hasn't talked about L' Hôpital's Rule up to that point!

Thank you for your help in advance...

Last edited by skipjack; June 28th, 2017 at 06:26 AM.
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June 28th, 2017, 06:19 AM   #2
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Hint: multiply numerator and denominator by cos(x)((1+tan(x))^0.5+(1+sin(x))^0.5).
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June 28th, 2017, 06:29 AM   #3
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$\displaystyle x\rightarrow 0$
$\displaystyle \lim \frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin x}}{x^3}=\lim \frac{\tan{x}
- \sin x}{2x^3}=\frac{1}{2} \lim \frac{\sin x(\frac{1}{\cos x}-1)}{x^3}=$
$\displaystyle =\frac{1}{2} \lim \frac{(\frac{1}{\cos x}-1)}{x^2}\cdot \lim \frac{\sin x}{x}=\frac{1}{2} \lim \frac{1-\cos x}{x^2 \cos x}=\frac{1}{2} \lim \frac{1-\cos x}{x^2}=\frac{1}{4} $
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Last edited by skipjack; June 28th, 2017 at 08:25 AM.
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June 28th, 2017, 06:34 AM   #4
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thanks, but I still don't know what to do with the x^3 that's left in the denominator..
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June 28th, 2017, 06:37 AM   #5
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I got it! thanks a lot idontknow and skipjack
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June 28th, 2017, 08:31 AM   #6
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(1 - cos(x))/x² = 2sin²(x/2)/(4(x/2)²) = (1/2)(sin(x/2)/(x/2))²
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