Thursday

March 5, 2015

March 5, 2015

Posted by **Matt** on Monday, March 5, 2007 at 5:25pm.

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)/(1-x^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.

**Answer this Question**

**Related Questions**

Math-Limits - sqrt(1+tan x)-sqrt(1+sin x) lim all divided by x^3 x-->0

Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...

math calculus please help! - l = lim as x approaches 0 of x/(the square root of...

Math/Calculus - Solve the initial-value problem. Am I using the wrong value for ...

Math Help please!! - Could someone show me how to solve these problems step by ...

Math(Roots) - sqrt(24) *I don't really get this stuff.Can somebody please help ...

Calculus - Second Order Differential Equations - Posted by COFFEE on Monday, ...

Inequality - When I solve the inquality 2x^2 - 6 < 0, I get x < + or - ...

Trig - Find the exact values of the six trigonometric functions 0 if the ...

Calculus - Second Order Differential Equations - Solve the initial-value problem...