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

• UrbanXrisis
In summary, Data rearranged the limit to a different form in his first response. He then said that the derivative is already solved with the derivative's definition.
UrbanXrisis
$$lim_{h->0}\frac{ln(2+h)-ln2}{h}$$

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.

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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?

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.

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!

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.

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.

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 .

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

Daniel.

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 don't 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

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.

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

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

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

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dextercioby said:
Folklore says that Leibniz discovered diff.calculus while playing with tangents to curves...
Daniel.
Quite right but i was referring to the method that is used, it was first proposed before calculus by Fermat to find the slope of the line tangent to a point, although it's not the same notation as in calculus here's http://math.kennesaw.edu/~jdoto/13.pdf (goto Fermat's method) .

Last edited by a moderator:

1. What is the 'h' on the denominator?

The 'h' on the denominator refers to the variable used in the denominator of a mathematical expression.

2. Why do we need to get rid of the 'h' on the denominator?

In certain mathematical operations, having the variable 'h' in the denominator can make it difficult to solve the equation or find the limit. Removing the 'h' can simplify the problem and make it easier to solve.

3. How do we get rid of the 'h' on the denominator?

The process of removing the 'h' on the denominator depends on the specific mathematical expression or problem. It may involve factoring, simplifying, or using algebraic manipulation to eliminate the 'h' term.

4. Are there any limitations to getting rid of the 'h' on the denominator?

Yes, there are certain situations where it may not be possible or appropriate to remove the 'h' on the denominator. This could be due to the structure of the equation or the mathematical properties involved.

5. Can getting rid of the 'h' on the denominator change the solution to a problem?

Yes, removing the 'h' on the denominator can change the solution to a problem. It is important to carefully consider the steps taken to remove the 'h' and ensure that the solution is still accurate and valid.

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