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Mathematics 17 Online
OpenStudy (anonymous):

Calculus study guide. HELP? I need these questions answered so I can study this for the Final Exam.

OpenStudy (anonymous):

I'll try. So for Oscilliating Limit. You have a value that is bouncing between two numbers. In the case of \[\lim_{x \rightarrow 0} \sin \frac{ 1 }{ x }\] It bounces very rapidly between -1 and 1. So the limit does not exist.

OpenStudy (anonymous):

This is because the limit isn't headed toward a single value. Which others did you have?

OpenStudy (anonymous):

I have a sheet......

OpenStudy (anonymous):

Cool, I'll try my best. There are so many things I'm trying to get a handle on as well.

OpenStudy (anonymous):

You don't have to answer them all but any that you can will be appreciated!!!

OpenStudy (anonymous):

Okay, after looking up ivt for the last 15mins I can tell you that my book sucks at explaining it. I'll keep pushing on the different questions.

OpenStudy (anonymous):

okay ty.

OpenStudy (anonymous):

Sorry, I am rather useless at this. I'm currently going over this at khan, because I am trying to get this under raps by Monday. https://www.khanacademy.org/math/calculus/differential-calculus/der_common_functions/v/proof--d-dx-ln-x----1-x

OpenStudy (anonymous):

Going for the lim x->0 (ln(x+h)-lnx)/h, I was actually trying to say that (lnx +ln h -lnx) / h while working out the problem. Stupid addtion rules.

OpenStudy (anonymous):

Okay, I think I've got it. In order to prove lnx=1/x \[\lim_{h \rightarrow 0} \frac{ \ln(x+h)-lnx }{ h }\] all you need to know is: \[\frac{ 1 }{ h } = \frac{ 1 }{ u }\] \[u=\frac{ h }{ x } \]\[xu=h\] \[\log \frac{ a }{ b } = \log a - \log b\] \[a \log b = \log b^a\]\[a ^{bc} = a ^{b ^{c}} \] and last but not least \[e= \lim_{u \rightarrow 0} (1+u)^{\frac{ 1 }{ u }}\]

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