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so I need to get rid of the 'h' on the denominator, but how can I do that?

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- Thread starter UrbanXrisis
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- #1

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so I need to get rid of the 'h' on the denominator, but how can I do that?

- #2

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[tex]\frac{\ln (2+h) - \ln(2)}{h} = \ln \left[\left( 1 + \frac{h}{2}\right)^{\frac{1}{h}}\right][/tex]

Do you see why?

Now, this may or may not help you, depending on how much you know. If you know

[tex] \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^n = e^x,[/tex]

and

[tex] \lim_{x \rightarrow a} \ln f(x) = \ln \left( \lim_{x \rightarrow a} f(x) \right)[/tex]

then this is enough for you to do the question.

Do you see why?

Now, this may or may not help you, depending on how much you know. If you know

[tex] \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^n = e^x,[/tex]

and

[tex] \lim_{x \rightarrow a} \ln f(x) = \ln \left( \lim_{x \rightarrow a} f(x) \right)[/tex]

then this is enough for you to do the question.

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- #3

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my book explains it as... [tex]lim_{h->0}\frac{ln(2+h)-ln2}{h}=f'(2)[/tex], where [tex]f(x)=lnx[/tex].

what does that mean?

- #4

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It can't get any more clear/elementary than that.

Daniel.

- #5

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The derivative of a function [itex]f(x)[/itex] at the point [itex]x[/itex] with respect to [itex]x[/itex] is defined to be

[tex]f^\prime (x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}[/tex]

if this limit exists.

The numerator of the limitand, [itex]f(x+h)-f(x)[/itex], is the change in [itex]f[/itex] between [itex]x[/itex] and [itex]x+h[/itex]. The denominator is the change in [itex]x[/itex], which is just [itex]h[/itex]. As the change in [itex]x[/itex] approaches zero (ie. when we take the limit), we get the

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Folklore says that Leibniz discovered diff.calculus while playing with tangents to curves...

Daniel.

Daniel.

- #8

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okay,I understand all that you have said. f(x) is ln(x).

how do I actually solve the problem?

how do I actually solve the problem?

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just to play devils advocate ...dextercioby said:Folklore says that Leibniz discovered diff.calculus while playing with tangents to curves...

Daniel.

Sure, after he stole the idea from Newton! :tongue2:

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Daniel.

- #11

SpaceTiger

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UrbanXrisis said:how do I actually solve the problem?

See how Data rearranged the limit to a different form in his first response? First, verify that you understand how he got it to that form. Then, see if you can use his second and third formulae to get a result.

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UrbanXrisis said:okay,I understand all that you have said. f(x) is ln(x).

how do I actually solve the problem?

What do you mean...?It's already solved with the derivative's definition.

Daniel.

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- #14

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Not to mention in different context.Was was interested in maths,the other in physics...

Daniel.

Daniel.

- #15

SpaceTiger

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dextercioby said:What do you mean...?It's already solved with the derivative's definition.

If this is like the structure of my old calc class, he may not have covered derivatives yet.

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I dont see have Data's second and third equation fit into all of this

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SpaceTiger said:If this is like the structure of my old calc class, he may not have covered derivatives yet.

I've learned derivatives, I think I just forget the definition of a derivative and how to solve the definition, it was a while back and my class is doing some basic review

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[tex] \lim_{h\rightarrow 0} \frac{\ln(2+h)-\ln 2}{h}=:\left \left(\frac{d}{dx}\ln x\right) \right|_{x=2} =\left\frac{1}{x}\right|_{x=2}=\frac{1}{2} [/tex]

Daniel.

- #19

SpaceTiger

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UrbanXrisis said:I've learned derivatives, I think I just forget the definition of a derivative and how to solve the definition, it was a while back and my class is doing some basic review

Well, if you're allowed to solve this problem by invoking the derivative of a logarithm, just do that. Otherwise:

[tex]\frac{\ln (2+h) - \ln(2)}{h} = \ln \left[\left( 1 +\frac{h}{2}\right)^{\frac{1}{h}}\right][/tex]

Explain to me how this is true so that I can be sure that you're following.

- #20

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oh my, okay, thanks, I get it

- #21

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SpaceTiger said:Well, if you're allowed to solve this problem by invoking the derivative of a logarithm, just do that. Otherwise:

[tex]\frac{\ln (2+h) - \ln(2)}{h} = \ln \left[\left( 1 +\frac{h}{2}\right)^{\frac{1}{h}}\right][/tex]

Explain to me how this is true so that I can be sure that you're following.

[tex]\frac{\ln (2+h) - \ln(2)}{h} = \frac{ln(\frac{2+h}{h})}{h}= \ln \left[\left( 1 +\frac{h}{2}\right)^{\frac{1}{h}}\right][/tex]

thanks but I understood what dextercioby did :)

- #22

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[tex]\frac{\ln (2+h) - \ln(2)}{h} = \frac{\ln(\frac{2+h}{h})}{h}= \ln \left[\left( 1 +\frac{h}{2}\right)^{\frac{1}{h}}\right][/tex]

there's a problem in your middle step (and thus the result does not follow from your middle step). Do you see where?~

- #23

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foolish mistake

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however, I did not know that [tex]\frac{ln(x)}{z}=ln(x)^{\frac{1}{z}}[/tex]

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indeed, since

[tex]x = \ln e^x[/tex]

and thus for

[tex]e^{\alpha \ln \gamma} = \gamma^\alpha[/tex]

we get

[tex]\ln \gamma^\alpha = \ln e^{\alpha \ln \gamma} = \alpha \ln \gamma.[/tex]

or

[tex] \ln \gamma^\alpha = \alpha \ln \gamma[/tex]

Note that you should be a little careful: what you put in your last post is not*quite* right. It is not

[tex]\frac{\ln x}{z} = \ln (x)^{\frac{1}{z}}[/tex]

but instead

[tex]\frac{\ln x}{z} = \ln \left(x^{\frac{1}{z}}\right).[/tex]

[tex]x = \ln e^x[/tex]

and thus for

[tex]e^{\alpha \ln \gamma} = \gamma^\alpha[/tex]

we get

[tex]\ln \gamma^\alpha = \ln e^{\alpha \ln \gamma} = \alpha \ln \gamma.[/tex]

or

[tex] \ln \gamma^\alpha = \alpha \ln \gamma[/tex]

Note that you should be a little careful: what you put in your last post is not

[tex]\frac{\ln x}{z} = \ln (x)^{\frac{1}{z}}[/tex]

but instead

[tex]\frac{\ln x}{z} = \ln \left(x^{\frac{1}{z}}\right).[/tex]

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