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Calculus/L'Hopital's rule doesnt work

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Expert: - 12/2/2009


Question
My professor gave my class as interesting question today.  He told us to go find the answer but after about an hour of searching online and in my textbook i cant figure it out.

Question: "Explain why we may NOT use L'Hopital's Rule to prove lim x->0 of sin(x) / x."

I understand you can do it using the rule, but my professor says it doesn't prove that it is 1, it only reinforces the geometric prove.  The thing i don't understand is why.

Answer
That statement is wrong.  L'hospital's rule does work.

We have f(x) = g(x)/h(x).
It is seen that g(x) = sin(x), so g'(x) = cos(x).
It is seen that h(x) = x, so h'(x) = 1.

It is known that lim(x->0) of cos(x) = 1 and the lim(x->0) of 1 = 1, so the answer is 1/1 = 1.

This is true since the Taylor's Expansion of sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ....
and if that is divided by x, that expansion be comes 1 - x^2/3! + x^4/5! - x^6/7! +....
which is 1 as x goes to 0.

This is because all of the terms ((-1)^n)x^(n-1)/n! go to 0
except for the 1st term in which n = 0.

Calculus

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