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November 23, 2014

November 23, 2014

Posted by **Matt** on Tuesday, March 6, 2007 at 6:41pm.

If not, could you try to answer this tough, for me, question. Thanks a lot.

lim

x-->0

Square root (1 + tan x) - Square root (1 + sin x) all over x^3.

Limits involving analytical functions are most easily calculated using series expansion techniques. In this case we want to calulate a limit for x--> 0, so the natural thing to do is to examine how the functions that are involved in this limit behave in the vicinity of x = 0.

In this problem you need to use that:

sin(x) = x - x^3/6 + term of order x^5

cos(x) = 1 - x^2/2 + term of order x^4

You find tan(x) by dividing sin(x) by cos(x).

Both tan(x) and sin(x) tend to zero for x -->0 so you need to know how the square root function behaves near 1. For y close to zero we have:

sqrt[1 + y] = 1 + 1/2 y + (1/2)(-1/2)/2 y^2 + (1/2)(-1/2)(-3/2)/6 y^3+ term of order y^4

You just insert the series expansion for the tan and sin in y and subrtact the two terms...

It is possible to calculate limits without using series expansions. But that's more tedious because ultimately you must make use of the way the fiunctions behave near zero. If you don't make use of the series expansions then the information contained in them must be derived in some form during that derivation.

Let me give some other examples:

Lim x-->0 sin(x)/x = 1

If you use that sin(x) = x + term of order x^3, then you immediately see that:

sin(x)/x = 1 + term of order x^2

Take limit x-->0 of both sides and you find the desired result.

Lim x-->0 (sin(x)-x)/x^3 = -1/6

Use that:

sin(x) - x = - x^3/6 + term of order x^5

Divide both sides by x^3:

(sin(x) - x)/x^3 =

- 1/6 + term of order x^2

Take the limit x-->0 to find te desired result.

You can also calculate the latter limit by applying L'Hopital's rule three times! :)

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