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# Calculus/Limits and l'Hospital's rule.

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Question

maths question
How can I do question number in the following picture.

Answer
I assume you mean No 8.

This is going to be tricky, and I only have the first part for you.  Perhaps you can use it to push the second part through.

Let  y = (sin x/x)^1/x

ln y = 1/x[ln sin x - ln x]   ==> If we find lim ln y, lim y = exp(lim ln y)

ln y = [ln sin x - ln x]/x

Diff(t&b):(l'Hospital's rule, to be repeated as many times as needed)

cos x/sin x - 1/x
-------------------
1

x cot x - 1
------------
x

Diff(t&b):(again)

x(-csc^2 x) + cot x

-x(1/ sin^2) + cos/sin

-x + sin cos
--------------
sin^2

Diff(t&b):
-1 + cos^2 - sin^2
------------------
2 sin cos

-1 + cos(2x)
--------------
sin(2x)

Diff(t&b):

-2 sin(2x)
-----------
2 cos(2x)

- sin(2x)
-----------
cos(2x)

As x -> 0, this -> 0/1 = 0.

So lim[x->0] ln y = 0

And so lim[x->0] y = 1.

Calculus

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#### Paul Klarreich

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