MHB Differentiation With L'Hopital's Rule

[ardentmed]

Hey guys,

Need some more help again. I'll keep it brief.

This thread is only for question 2ab. Please ignore question 1:

For 2a, I simply employed L'Hopital's Rule since 0/0 is indeterminate form. Thus, my final answer came out to be: ln6-ln3.

As for 2b, I computed an indeterminate form in the form of 0^0, which requires L'Hopital's Rule.

After taking the derivative of both the numerator and denominator respectively, I ultimately obtained 0.
Thanks in advance.

 

[Sudharaka]

Hey guys,

Need some more help again. I'll keep it brief.

This thread is only for question 2ab. Please ignore question 1:For 2a, I simply employed L'Hopital's Rule since 0/0 is indeterminate form. Thus, my final answer came out to be: ln6-ln3.

As for 2b, I computed an indeterminate form in the form of 0^0, which requires L'Hopital's Rule.

After taking the derivative of both the numerator and denominator respectively, I ultimately obtained 0.
Thanks in advance.

Hi ardentmed, :)

The answer for 2a is correct. However you can simplify it a little bit using the rules of logarithms. Could you see how?

The answer for 2b is incorrect. Could you write down your attempt at solving it so that I can tell where you made the mistake. :)

 

[ardentmed]

Hi ardentmed, :)

The answer for 2a is correct. However you can simplify it a little bit using the rules of logarithms. Could you see how?

The answer for 2b is incorrect. Could you write down your attempt at solving it so that I can tell where you made the mistake. :)

Sure, I carried down the x as I converted the function to its logarithmic form, giving me xlntan(2x), I then proceeded to move the x down into the denominator (so it becomes 1/x in the denominator).

Moreover, I applied L'Hopital's Rule and took the derivative (albeit I'm not too sure if the function actually gives 0/0, but it looks like it).

This gave me:

2sec^2(2x) / tan(2x) * (-x^2 / 1) = infinity.

So the answer would be e^infinity = +infinity.

I'm at a loss as to how I arrived at this answer, as I'm almost definite that it is incorrect. I'd appreciate some help.

Thanks in advance.

 

[Dethrone]

Have you tried to apply L'hopital's rule again? 2sec^2(2x) / tan(2x) * (-x^2 / 1) seems to be in that form.

 

[ardentmed]

Have you tried to apply L'hopital's rule again? 2sec^2(2x) / tan(2x) * (-x^2 / 1) seems to be in that form.

Oh, is it in indeterminate form? I computed it to be infinity, but it could potentially be 0/0, but I'm not too sure.

 

[Dethrone]

$$\ln\left({y}\right)= \lim_{{x}\to{0^+}} \frac{2(sec2x)^2(-x^2)}{tan2x} $$ is of the form 0/0.

 

[ardentmed]

$$\ln\left({y}\right)= \lim_{{x}\to{0^+}} \frac{2(sec2x)^2(-x^2)}{tan2x} $$ is of the form 0/0.

Alright, so I performed another round of L'Hopital's rule and computed lny=0 because:

lny= $\lim_{{x}\to{0+}} (-4x^2 * tan2x-x)/1 $

Ergo,

y= e^0

Thus, the limit is 1.

Am I on the right track? Thanks.

 

[Dethrone]

That is correct, the limit evaluates to 1.