L'Hospitals Rule for finding the limits of a function

[KiNGGeexD]

Above is a function which I have been given to take the limits of, now we were briefly introduced to L'Hospitals rule and given this example!

Here is what I have done so far...l..

Basically I know that I have to take the derivatives of each of the functions in the quotient and then tend it to the limits, but I'm having a little trouble!

What I've done so far I took from reading elsewhere (textbooks) for dealing with products and the rest was just guess work as to how to solve the problem we literally scrapped the surface of this and it took no longer than 5 minutes to do so! I hope this is enough information

Any help would be appreciated

Regards

 

[KiNGGeexD]

Ahh I've just realized that it would be better to take ln(x) as my function g(x)

To give infinity/infinity Otherwise (the initial way) would be a combination of zero and infinity, my intuition on this is not so good as I said, I may have scared potential help away but it's the truth

 

[KiNGGeexD]

Attached is my attempt at the solution

 

[dauto]

The 1st page of your last post seems correct. The second page isn't. (Hard to learn calculus when you can't remember how to add fractions...)

 

[KiNGGeexD]

Sorry for the number of posts but is that to my initial post or revised one:)

 

[KiNGGeexD]

Ahh I see what I've done, I put the denominator of the fraction of the numerator as the denominator of the whole fraction! Whoops!

 

[Mark44]

This is strictly a calculus problem, so I'm moving it to the Calculus & Beyond section, which is where it should have been posted.

 

[KiNGGeexD]

Ok, thanks! It was out of habit that I posted it here, apologies

 

[KiNGGeexD]

Dauto, would this be correct?

 

[KiNGGeexD]

Never mind I have solved this problem, my derivative for g(x) was incorrect and once I amended this and added the fractions properly I ended up with

-2/π Which I'm pretty sure is the correct answer

 

[Curious3141]

Never mind I have solved this problem, my derivative for g(x) was incorrect and once I amended this and added the fractions properly I ended up with

-2/π


Which I'm pretty sure is the correct answer

Correct answer.

Quite easy to do. Rearrange that to ##\displaystyle \frac{\ln x}{\frac{1}{tan(\frac{\pi x}{2})}}## which is a 0/0 limit. The top is trivial to differentiate. The bottom is a quick application of Chain Rule and a bit of simplification. The result quickly follows.

 

[KiNGGeexD]

Brilliant thanks! Typically stupid errors which trip me up! I should be more careful

 

[Ray Vickson]

Brilliant thanks! Typically stupid errors which trip me up! I should be more careful

Another way (almost equivalent to l'Hospital's method, but sometimes much easier): just expand both ##\ln(x)## and ##\tan(\pi x/2)## in series about ##x=1##, essentially by setting ##x = 1+t## and expanding in powers of ##t##. Of course, ##\ln(1+t) = a_1 t + a_2 t^2 + \cdots## while ##\tan(\pi(1+t)/2) = b_0/t + b_1 t + b_2 t^2 + \cdots##. As ##t \to 0## their product approaches ##b_0 a_1##, so you need only retain the first terms in both expansions.

 

[KiNGGeexD]

We haven't done much on Taylor expansion, I know the generic formulae and how to calculate it for functions but as for taking limits with it we haven't went over it much! Although we were assigned a question similar to this with Taylor expansion! You will probably see it up here in a few hours haha