Question

LIMITS. help?

Answer 1

YES?

Answer 2

what?

Answer 3

sorry slow net connection :|

Answer 4

HELP?

Answer 5

can someoone explain why the answer is 0 ????

Answer 6

i don't get it :(

Answer 7

This is \(\log(x)\) base 10, right?

Answer 8

yup

Answer 9

Try evaluating \(\frac{\cos(x)}{\log(x)}\) at \(x=-1\), \(x=-0.1\), \(x=-0.001\), \(x=-0.0001\),...

Do you see a pattern?

Answer 10

limit(0+)log(x)=infinite
limit(0+)cos(x)=1
1/infinite=0 ;)

Answer 11

sorry about the other answer, i wrote +oo while it's 0+

Answer 12

limitless negative infinity?

Answer 13

Err.. Why would it be negative infinity?

Answer 14

yess negative
check this http://www.wolframalpha.com/input/?i=lim%280%2B%29logx

Answer 15

Negative infinity is incorrect.

Answer 16

dont know the pattern :|

Answer 17

@kenji07 got it or not?
@Limitless it is negative infinity

Answer 18

This is the result:
http://www.wolframalpha.com/input/?i=%5Clim_%7Bx+%5Cto+0%5E%7B%2B%7D%7D+%5Cfrac%7B%5Ccos%28x%29%7D%7B%5Clog_%7B10%7D%28x%29%7D

Answer 19

There is a way to explain this that involves L'Hopital's rule, I think. I'm not certain. Let me think.

Answer 20

i dont get it. i know the answer is zero but why zero?

Answer 21

as i told you
limit(0+)log(x)= -infinite
limit(0+)cos(x)=1
1/infinite=0 ;)

Answer 22

@mzbc, you are so incorrect it is becoming horrid. Please do not continue to use incorrect statements and confuse kenji.

Answer 23

no it is correct, and for sure

Answer 24

i really dont understand. its just that we just started learning about this

Answer 25

Ok. Here we go, an attempt at an explanation:

\[\lim_{x \to 0^{+}}\frac{\cos(x)}{\log_{10}(x)}=\frac{\cos(0)}{\log_{10}(0)}=\frac{1}{-\infty}.\]

This is considered an "indeterminate form". L'Hopital's rule exists to evaluate these limits. There is absolutely no such thing as the expression \(\frac{1}{-\infty}\) in standard analysis.

Since this form is indetermine, we can apply L'Hopital's rule which states:
\[\lim_{x \to a}\frac{f(x)}{g(x)}=\lim_{x \to a}\frac{f'(x)}{g'(x)}.\]
\(\cos'(x)=-\sin(x)\) and \(\log_{10}'(x)=\frac{1}{x\ln(x)}.\)

Therefore, we get \[\lim_{x \to 0^{+}}\frac{\cos(x)}{\log_{10}(x)}=\lim_{x \to 0^{+}}\frac{-\sin(x)}{\frac{1}{x \ln(x)}}=\lim_{x \to 0^{+}}-\sin(x)x\ln(x)=0.\]

We see this from the following graph:
http://www.wolframalpha.com/input/?i=-sin%28x%29x*ln%28x%29+at+x%3D-0.1+to+x%3D-0.00001

The blue line (i.e. \(-sin(x)x\ln(x)\)) is approaching \(0\) as we approach \(0\) from the right.

Answer 26

One mistake, sorry!
\(\log_{10}'(x)=\frac{1}{x\ln(10)}\)

Therefore, we should have:

\[\lim_{x \to 0^{+}}\frac{\cos(x)}{\log_{10}(x)}=\lim_{x \to 0^+}-\sin(x)x\ln(10)=0.\]

Sorry about that! Forget the bottom of the post; it's technically not correct.

Answer 27

THANKS for explaining ! im now analyzing it.

Answer 28

I feel like I need to present an explanation of why Wolfram Alpha says that \(\frac{1}{\infty}=0.\)

Here is such explanation: In what's called the extended real numbers, the expression \(\frac{1}{\infty}\) has meaning. It is defined to be \(0.\) However, in the normal real numbers, \(\frac{1}{\infty}\) is completely undefined. There is no such meaning for this expression.

While the extended real numbers does give the same answer as the typical real numbers, I don't think it's fair to assume that we are working in the extended real numbers. This is why I felt mzbc's reasoning was flawed. There was no reasonable assumption that we were working in the extended real numbers.

Here is Wiki's article explaining them:
http://en.wikipedia.org/wiki/Extended_real_number_line

Answer 29

ok. thanks a lot for the help ! :D

Answer 30

You're welcome! :D