Question

Please refer to the image! Calculus

Answer 1

Thanks in advance for any help!

Answer 2

you know L'Hospitals rule?

Answer 3

yeah, but it did not work for me...

Answer 4

our prof told us to convert this limit to derivative form and solve it.

Answer 5

wolfram told me this answer 3/10, but i dont know to get it

Answer 6

yeah I know the answer as well.

Answer 7

I am wondering about the process..

Answer 8

Do you understand what it means to convert the limit to derivative form?

Answer 9

Because when you do it it gets really easy.

Answer 10

I do not know...

Answer 11

any help would be greatly appreciated!!

Answer 12

Ok the definition of derivative is:\[\frac{df}{dt}(t)=\lim_{h \rightarrow 0}\frac{f(t+h)-f(t)}{h}\], you need to make the limit get in the same form of this equation and of course you'll need to change variables at some point because in one the limit is to the infinity and on the other to zero.
Try it.

Answer 13

ok, this is past exam for my upcoming exam. I have learned two equations for the definition of derivative which I am fully aware. The equation you have given has h in the demoninator. My limit does not have denominator ...

Answer 14

I was trying to use the other one lim x approaches a

Answer 15

f(x)-f(a)/x-a

Answer 16

The other definition of derivative. I am stuck in making progress.

Answer 17

if you could tell me, the derivative form of this limit problem, I will go ahead and solve it for sure

Answer 18

Its the same thing, I just called h=x-a, and when x-->a, h-->0, its just more compact and easier to work with.

Answer 19

I think the first eq is equal to f prime (x) and second equation is equal to f prime a. First one gives an equation for derivative and second one gives a value for the derivative.

Answer 20

No, wait, to do this you need to see wich part of the equation relates with wich part of the derivative formula. The hole equation transforms into the derivative formula, not by parts.
In both you have somthing minus something else, try to go from there and don't forget that the derivative is taken at a specific point.

Answer 21

Just a hint, you need to have f(x) on both terms, and in one of them you need to have something that goes to zero.

Answer 22

thank you for your help. The part you are asking for me to do is the reason why I am posting here for help. Anyway, I appreciate your help for sure!

Answer 23

If you want I can show it for you. I was just trying to make you see it for yourself.

Answer 24

\[\lim_{x \rightarrow \infty}\sqrt[10]{x^10+3x^9}-x\]\[\lim_{x \rightarrow \infty}\sqrt[10]{x^{10}\left(1+\frac{3}{x} \right)}-x\]\[\lim_{x \rightarrow \infty}x \sqrt[10]{1+\frac{3}{x}}-x\]\[\lim_{x \rightarrow \infty}x\left(\sqrt[10]{1+\frac{3}{x}}-1\right)\]Now you see that it is starting to get similar, both have 1, and 3/x tends to 0, but we are still missing the denominator.
I define 3/x as h and do the acording substitutions, multiplicating everything by h/h=1 and therefore does not change the function.
\[\frac{3}{x}=\Delta x\]\[3=x \Delta x\]\[\lim_{\Delta x \rightarrow 0}\frac{\Delta x}{\Delta x}x\left(\sqrt[10]{1+\Delta x}-1\right)\]Since 3=x Delta x\[3\lim_{\Delta x \rightarrow 0}\frac{\sqrt[10]{1+\Delta x}-\sqrt[10]{1}}{\Delta x}=3\frac{d}{dx}\left(\sqrt[10]{x}\right)(1)=\frac{3}{10}\frac{1}{\sqrt[10]{1^9}}=\frac{3}{10}\]
If you have any doubts just say it.

Answer 25

Ok, I see now. It makes sense. I see how the eq becomes the definition equation.

Answer 26

Really appreciate it!

Answer 27

@ivanmlerner
I hope you don't mind me asking. How did you get this: \[3\frac{d}{dx}\left(\sqrt[10]{x}\right)(1)=\frac{3}{10}\frac{1}{\sqrt[10]{1^9}}?\]
I know \[3\frac{d}{dx}\left(\sqrt[10]{x}\right)(1)=\frac{3}{10}\frac{1}{\sqrt[10]{x^9}}\]But I don't understand why you put x=1 there.

Answer 28

The definition of the derivative is:\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]When you compare it with the equation I found in the first formula of the last line, you see that you are taking the derivative at x=1 of the function c^(1/10)

Answer 29

Ahh!!! Got it! Thanks!!!