Anyone spot the problem, simple limits question

marcus2001

I have that:

\displaystyle\lim_{x\to 0}\dfrac{\sqrt{9-x} - 3}{x} = \displaystyle\lim_{x\to 0}\dfrac{\sqrt{9-x} - \sqrt{9}}{x} = \displaystyle\lim_{x\to 0}\dfrac{\frac{(9-x)-9}{\sqrt{9-x} + \sqrt{9}}}{x} = \displaystyle\lim_{x\to 0}\dfrac{-1}{\sqrt{9-x} + 3} =-\frac{1}{6}

When in fact the limit is -1/2 when done expressing

\sqrt{9-x}

as a Taylor series..

EDIT: (Where the third stage follows from the identity

a - b = \dfrac{a^2 - b^2}{a+b}

)

(edited 10 years ago)

Reply 1

marcus2001

I understand how to do it expanding as a Taylor expansion, I just don't get why the above doesn't work

Reply 2

marcus2001

Ffs never mind, watched an extra minute of the lecture and realised the lecturer had made a mistake, the answer is -1/6

What a waste of time typing out that colossus in latex

Reply 3

Carpetman

d/dx((9-x)^{(1/2)})[br][br][br]= (-{1/2})(9-x)^{(-1/2)}

as x tends to 0

= -(1/3)(1/2) = -(1/6)

(edited 10 years ago)

Reply 4

marcus2001

Original post by Carpetman

d/dx =(9-x)^{(1/2)} [br][br][br]= (-{1/2})(9-x)^{(-1/2)}

as x tends to 0

= -(1/3)(1/2) = -(1/6)

Yeah I should have just checked using l'hopitals instead of wasting my time trying to find the error here

Reply 5

FireGarden

Your problem is quite clear.

-\dfrac{1}{6}

is the right answer.

Reply 6

Carpetman

Original post by marcus2001

Yeah I should have just checked using l'hopitals instead of wasting my time trying to find the error here

Nothing worse than a mistake in the notes, I've spent hours on problems trying to find out where I've gone wrong only to find out the lecturer made a simple error and I was in fact right!

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Anyone spot the problem, simple limits question

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