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Differentiable function q

I can't even do 1i. I first considered when x_0 is real but never equal to 1. Then I considered the limit of differentiable function definition. But this is a very dirty limit, my guess is that it doesn't exist but how would I prove this?

I am aware that u can check the lh limit and rh limit then conclude they are different but this limit is too dirty.

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(edited 9 years ago)

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Reply 1

cooldudeman

This is what I have now. Im guessing it is missing so much justification but is this on the right lines?

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Reply 2

DFranklin

Original post by cooldudeman

This is what I have now. Im guessing it is missing so much justification but is this on the right lines?No. Looks like you've completely forgotten what a limit is, to be honest.

Reply 3

cooldudeman

Original post by DFranklin

No. Looks like you've completely forgotten what a limit is, to be honest.

Can you start me off please

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Reply 4

TheIrrational

Original post by cooldudeman

Can you start me off please

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In the second post you posted, can you explain why the 3rd line is

+\infty

? I don't imagine it is.

Reply 5

cooldudeman

Original post by TheIrrational

In the second post you posted, can you explain why the 3rd line is

+\infty

? I don't imagine it is.

well I really don't know what to do with the top part of the fraction and the h is gonna approach 0 from the right so its going to be positive.
I don't know what do at all tbh

Reply 6

cooldudeman

Original post by DFranklin

No. Looks like you've completely forgotten what a limit is, to be honest.

I have redone the first question. Could you have a look please and tell me if I have done it correctly. I am a bit unsure on 1ii

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Reply 7

DFranklin

Original post by cooldudeman

I have redone the first question. Could you have a look please and tell me if I have done it correctly. I am a bit unsure on 1iiMuch better, but

\displaystyle \lim_{x \to 1} \cos\left(\dfrac{1}{x-1}\right)

doesn't exist (and neither does the similar sin limit), so the line where you take those limits is incorrect.

Spoiler

Reply 8

cooldudeman

Original post by DFranklin

Much better, but

\displaystyle \lim_{x \to 1} \cos\left(\dfrac{1}{x-1}\right)

doesn't exist (and neither does the similar sin limit), so the line where you take those limits is incorrect.

Spoiler

OK but after reading your spoiler, how are you supposed to phrase this last bit then? Since you said they don't have to exist.

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