What is the difference quotient of ln (x^3 -5)? Picture provided....

[Michael Santos]

Mentor note: Thread moved from technical math section, so is missing the homework template

.[ATTACH=full]234622[/ATTACH]

 

[BvU]

I wish you wouldn't write $\ln$ as $\operatorname {in}$

Please post in homework and tell us which relevant equations you think you will need to use ...

 

[Michael Santos]

I wish you wouldn't write $\ln$ as $\operatorname {in}$

Please post in homework and tell us which relevant equations you think you will need to use ...

How do i move this over to homework

 

[Mark44]

@Michael Santos, homework questions must be posted in one of the Homework & Coursework sections. I have moved your thread to the Calculus & Beyond section.

Do not put the problem details in the thread title. They should go in the problem description of the homework template.

All of the work you show in the image can be done using LaTeX. See our tutorial here -- https://www.physicsforums.com/help/latexhelp/.

For example, $$\lim_{h \to 0} \frac 1 h \ln\left(\frac{(x + h)^3 - 5}{x^3 - 5}\right)$$
Here's what I actually wrote: $$\lim_{h \to 0} \frac 1 h \ln\left(\frac{(x + h)^3 - 5}{x^3 - 5}\right)$$ All of this is described in our tutorial.

To complete the problem, I don't believe expanding the expression $(x + h)^3 - 5$ is any help. A better way to do this is to use L'Hopital's Rule.

Finally, there is no function named "In" -- the first letter is lowercase L, for logarithm.

 

[vela]

To complete the problem, I don't believe expanding the expression $(x+h)^3−5$ is any help. A better way to do this is to use L'Hopital's Rule.

Considering he's trying to calculate derivatives using the basic definition, I think L'Hopital's Rule isn't fair game.

After expanding $(x+h)^3$, you can rewrite the argument of the log in the form (1+something). Go from there.

 

[Michael Santos]

Considering he's trying to calculate derivatives using the basic definition, I think L'Hopital's Rule isn't fair game.

After expanding $(x+h)^3$, you can rewrite the argument of the log in the form (1+something). Go from there.

Yes the basic definition is what i am after.
Vela, will you be providing me with the rest of the answer?

 

[Mark44]

Yes the basic definition is what i am after.
Vela, will you be providing me with the rest of the answer?

No, that's against forum rules. @vela has given you a strong hint that involves polynomial division.

 

[Michael Santos]

Yes the basic definition is what i am after.
Vela, will you be providing me with the rest of the answer?

Vela here is what i have so far
$$\lim_{h\to 0} {\frac {\ln((x+h)^3 -5) - ln (x^3 - 5)} h} $$
$$\lim_{h\to 0}{\frac {\ln (x^3 +3x^2h +3xh^2 +h^3 -5) -ln (x^3 - 5)} h} $$
$$\lim_{h\to 0}{\frac {\ln (x^3 -5 +3x^2h +3xh^2 +h^3) -ln (x^3-5)} h} $$
I factored and used the product rule
$$\lim_{h\to 0}{{\frac {1} h} ×ln(x^3-5) +ln (1+{\frac {3x^2h+3xh^2+h^3} {x^3-5})} -ln (x^3-5)} $$
I can now eliminate the two $${ln (x^3-5)} $$ terms with subtraction and now i am stuck forever without you aid.

$$\lim_{h\to 0}{{\frac {1} h} ×ln(1+{\frac {3x^2h+3xh^2+h^3} {x^3-5})}}$$

 

[Michael Santos]

No, that's against forum rules. @vela has given you a strong hint that involves polynomial division.

Yes the basic definition is what i am after.
Vela, will you be providing me with the rest of the answer?

Mark here is what i have so far
$$\lim_{h\to 0} {\frac {\ln((x+h)^3 -5) - ln (x^3 - 5)} h} $$
$$\lim_{h\to 0}{\frac {\ln (x^3 +3x^2h +3xh^2 +h^3 -5) -ln (x^3 - 5)} h} $$
$$\lim_{h\to 0}{\frac {\ln (x^3 -5 +3x^2h +3xh^2 +h^3) -ln (x^3-5)} h} $$
I factored and used the product rule
$$\lim_{h\to 0}{{\frac {1} h} ×ln(x^3-5) +ln (1+{\frac {3x^2h+3xh^2+h^3} {x^3-5})} -ln (x^3-5)} $$
I can now eliminate the two $${ln (x^3-5)} $$ terms with subtraction and now i am stuck forever without you aid.
What now?
$$\lim_{h\to 0}{{\frac {1} h} ×ln(1+{\frac {3x^2h+3xh^2+h^3} {x^3-5})}}$$

 

[BvU]

Do you know a limit for $$\ln (1+\varepsilon)\over \varepsilon $$ when $\varepsilon\downarrow 0$ ?

 

[Michael Santos]

Do you know a limit for $$\ln (1+\varepsilon)\over \varepsilon $$ when $\varepsilon\downarrow 0$ ?

No i do not
It is approaching one

 

[BvU]

In that case things become difficult. What equations do you have in your toolkit that may help in this situation ? In the template they are called 'relevant equations'.

 

[Michael Santos]

In that case things become difficult. What equations do you have in your toolkit that may help in this situation ? In the template they are called 'relevant equations'.

Well i know it now, what can i do with it?

 

[BvU]

Well i know it now,

I suppose that refers to what you first said you did not know

No i do not

? Well, what do you get if you apply your new knowledge to your last expression in post #9 ?

 

[Ray Vickson]

Mark here is what i have so far
$$\lim_{h\to 0} {\frac {\ln((x+h)^3 -5) - ln (x^3 - 5)} h} $$

What now?
$$\lim_{h\to 0}{{\frac {1} h} ×ln(1+{\frac {3x^2h+3xh^2+h^3} {x^3-5})}}$$

You have
$$\frac{1}{h} \ln (1 + \underbrace{\frac{3 x^2 h + 3 x h^2 + h^3}{x^3-5}}_{H} ) $$
Here, $H \to 0$ whenever $h \to 0$, so you have $\ln(1+H)$ with small $|H|.$

 

[Michael Santos]

You have
$$\frac{1}{h} \ln (1 + \underbrace{\frac{3 x^2 h + 3 x h^2 + h^3}{x^3-5}}_{H} ) $$
Here, $H \to 0$ whenever $h \to 0$, so you have $\ln(1+H)$ with small $|H|.$

How doez that turn into the final answer of
$$\frac {3x^2} { \ x^3-5 \ }$$

 

[BvU]

Well i know it now,

You have $$\lim_{h\downarrow 0}\ {\ln (1+h )\over h} = 1$$ so what is $$\lim_{h\downarrow 0} \ {\ln (1+ \left(3x^2 \over x^3-5\right ) h )\over h} \quad \rm ? $$

 

[Ray Vickson]

How doez that turn into the final answer of
$$\frac {3x^2} { \ x^3-5 \ }$$

Plug it in and work it out for yourself. Remember that you want to take the limit as $h \to 0,$ so lots of complicating terms will go away.

 

[vela]

In another thread, you said you had solved a similar problem. I'm not sure why it's such a mystery now as to how to proceed with this one.

 

[Michael Santos]

Plug it in and work it out for yourself. Remember that you want to take the limit as $h \to 0,$ so lots of complicating terms will go away.

So
$$ \frac {3x^2h+3xh^2+h^3} { \ x^3-5 \ } $$ = h?

 

[Michael Santos]

You have $$\lim_{h\downarrow 0}\ {\ln (1+h )\over h} = 1$$ so what is $$\lim_{h\downarrow 0} \ {\ln (1+ \left(3x^2 \over x^3-5\right ) h )\over h} \quad \rm ? $$

-1.39...

 

[Michael Santos]

I figured it out thank you.

 

[BvU]

With @vela I'm still a bit worried. Any questions left over ? Or is everything now crystal clear and the next exercise won't be a problem any more ?