# Need to get rid of the 'h' on the denominator

$$lim_{h->0}\frac{ln(2+h)-ln2}{h}$$

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

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

Do you see why?

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

$$\lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^n = e^x,$$

and

$$\lim_{x \rightarrow a} \ln f(x) = \ln \left( \lim_{x \rightarrow a} f(x) \right)$$

then this is enough for you to do the question.

Last edited:
I really do not understand..

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

what does that mean?

dextercioby
Homework Helper
It means that the book applied the definition of the derivative of a function of one variable (which is $f(x)=\ln x$) at a point from its domain (which is $x=2$)...

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

Daniel.

It means that it is the derivative of the function $\ln{x}$ evaluated at $x=2$. A derivative can be thought of geometrically as a rate of change of one quantity with respect to another. In this case, it is the rate of change of the function $\ln x$ with respect to $x$, at the point $x=2$.

The derivative of a function $f(x)$ at the point $x$ with respect to $x$ is defined to be

$$f^\prime (x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$$

if this limit exists.

The numerator of the limitand, $f(x+h)-f(x)$, is the change in $f$ between $x$ and $x+h$. The denominator is the change in $x$, which is just $h$. As the change in $x$ approaches zero (ie. when we take the limit), we get the instantaneous rate of change of the function with respect to $x$.

Think of it as your good old rise over run to find the slope of a function. Slope is your rate of change, and Fermat(i think) came up with this neat way of computing the slope for a function that's not a line. You pick the point you want which is x then the difference between the point you want and any other point is h, so the denominator on the bottom is your run or difference in x, hence h on the bottom. Then you draw a secant line that crosses that function at the point x and another arbitrary point. The slope would then be (y-y1)/h, y1 is also f(x) in other notation and y is the function evaluated at your point x + your run h, or ((x+h),y), and in other notation f(x+h). Put it all together and you get your function, take the limit of that function as the run aproachers 0 and you get the derivative or slope at that point.

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

Daniel.

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

how do I actually solve the problem?

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

Daniel.
just to play devils advocate ...

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

dextercioby
Homework Helper
He didn't steal anything.Not even phylosophy and physics...He was a really brilliant man.In the same league with Newton,Euler & Gauss.

Daniel.

SpaceTiger
Staff Emeritus
Gold Member
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.

dextercioby
Homework Helper
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.

I'm purely joking, as far as I know, Leibniz and Newton discovered the idea independently of each other .

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

Daniel.

SpaceTiger
Staff Emeritus
Gold Member
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.

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

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

dextercioby
Homework Helper
If u did,then u'd understand that

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

Daniel.

SpaceTiger
Staff Emeritus
Gold Member
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:

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

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

oh my, okay, thanks, I get it

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

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

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

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

thanks but I understood what dextercioby did :)

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

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

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

foolish mistake

however, I did not know that $$\frac{ln(x)}{z}=ln(x)^{\frac{1}{z}}$$

indeed, since

$$x = \ln e^x$$

and thus for

$$e^{\alpha \ln \gamma} = \gamma^\alpha$$

we get

$$\ln \gamma^\alpha = \ln e^{\alpha \ln \gamma} = \alpha \ln \gamma.$$

or

$$\ln \gamma^\alpha = \alpha \ln \gamma$$

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

$$\frac{\ln x}{z} = \ln (x)^{\frac{1}{z}}$$

$$\frac{\ln x}{z} = \ln \left(x^{\frac{1}{z}}\right).$$