MathLimits
posted by Matt .
sqrt(1+tan x)sqrt(1+sin x)
lim all divided by x^3
x>0
Use that Sqrt[1+x] = 1+ 1/2 x + 1/2 (1/2)/2 x^2 + 1/2(1/2)(3/2)/6 x^3 +
O(x^4)
You can thus write the numerator as:
1/2 [tan(x)  sin(x)]  1/8 [tan^2(x)  sin^2(x)] + 1/16 [tan^3(x)  sin^3(x)] +
O(x^4)
Note that sin(x) and tan(x) are of order x near x = 0 so that the neglected terms are of order x^4 and won't contribute to the limit. The next step is to insert the series expansions of the sin and the tan. The series expansion of the tan can be obtained by division. We only need to work to order x^3, so:
tan(x) = sin(x)/cos(x) = (x  x^3/6)/(1x^2/2) + O(x^5) =
(x  x^3/6)(1 + x^2/2) + O(x^5) =
x + x^3/3 + O(x^5)
So, you see that in the first square brackets you get an term of order x^3 and you can divide by x^3 to get a finite limit for x > 0 (I find 1/12). The third square brackets yields a term of order x^5 and thus also doesn't contribute to the limit.
The term in the second square brackets is of order x^4 and thus yields zero when divided by x^3 and x >0.
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