Evaluate the following limit using L'Hopital's rule

[flaren5]

Hi, I was wondering if anyone has any insight on the following question...

Evaluate the following limit using L'Hopital's rule.

lim x(2tan^-1 x - p)
x→∞

= lim 1(2)[1/(1 + x²)](1)

...that is how I have started to solve it, I'm not sure if I'm on the right track. I know that the finally answer is -2, apparently.
But with the way things are working out I don't seem to be ending up with -2.

 

[mmm4444bot]

lim x(2tan^-1 x - p)
x→∞

= lim 1(2)[1/(1 + x²)](1)

Hi. I'm not sure what the second line above is supposed to equal, but let's remember that L’Hôpital's Rule works with indeterminant rational forms (eg: 0/0, ∞/∞, 0/∞, ∞/0).

In this exercise, evaluating the limit directly gives an indeterminant product form (∞*0).

So, let's first apply an algebraic manipulation, to rewrite the given product of functions as a ratio of functions.

Given a product of functions f(x)*g(x), we may rewrite the product as either of the following equivalent forms (with due consideration to division-by-zero issues):

f(x) ÷ 1/g(x)

g(x) ÷ 1/f(x)


In other words, we may rewrite the given expression as either of the following.


\(\displaystyle \dfrac{2arctan(x) - \pi}{1/x}\)


\(\displaystyle \frac{x}{\dfrac{1}{2arctan(x) - \pi}}\)



Try applying L’Hôpital's Rule to one of these rational forms.

If you need more help, please show your steps. Cheers :cool:

 

[flaren5]

Hi. I'm not sure what the second line above is supposed to equal, but let's remember that L’Hôpital's Rule works with indeterminant rational forms (eg: 0/0, ∞/∞, 0/∞, ∞/0).

In this exercise, evaluating the limit directly gives an indeterminant product form (∞*0).

So, let's first apply an algebraic manipulation, to rewrite the given product of functions as a ratio of functions.

Given a product of functions f(x)*g(x), we may rewrite the product as either of the following equivalent forms (with due consideration to division-by-zero issues):

f(x) ÷ 1/g(x)

g(x) ÷ 1/f(x)


In other words, we may rewrite the given expression as either of the following.


\(\displaystyle \dfrac{2arctan(x) - \pi}{1/x}\)


\(\displaystyle \frac{x}{\dfrac{1}{2arctan(x) - \pi}}\)



Try applying L’Hôpital's Rule to one of these rational forms.

If you need more help, please show your steps. Cheers :cool:

Thank you for you're help. Although, I'm tring to continue with L'Hopital's Rule, and seem to be having a hard time differentiating it. So far I have,


\(\displaystyle \dfrac{2arctan(x) - \pi}{1/x}\)

= lim 2arctan(x) - pi
x -1


2
= 1 + x²
-x^-2

I got this far, not sure if I'm on the correct path...but I seem to be having problems solving the problem in order to get an answer of -2.

 

[flaren5]

Thank you for you're help. Although, I'm tring to continue with L'Hopital's Rule, and seem to be having a hard time differentiating it. So far I have,


\(\displaystyle \dfrac{2arctan(x) - \pi}{1/x}\)

= lim 2arctan(x) - pi
x -1


2
= 1 + x²
-x^-2

I got this far, not sure if I'm on the correct path...but I seem to be having problems solving the problem in order to get an answer of -2.

I've tried several times to input my solution so that it can easily understood, but obviously I'm not sure how to type the math correctly. My apologies...if anyone can decipher what I'm trying to solve, it would be appreciated if you share the knowledge. Thank you.

 

[Deleted member 4993]

f = tan^-1(x)

df/dx = 1/(1 + x²)

\(\displaystyle \displaystyle \lim_{x \to \infty} \frac{2*tan^{-1}(x) - \pi}{\frac{1}{x}}\ \)

= \(\displaystyle \displaystyle \lim_{x \to \infty} \frac{\frac{2}{1 + x^2}}{-\frac{1}{x^2}}\ \)

= \(\displaystyle \displaystyle (-2) * \lim_{x \to \infty} \frac{x^2}{1 + x^2}\ \)

Now divide the numerator and the denominator by x² and evaluate the limit.

 

[mmm4444bot]

......2
= 1 + x²
.....-x^-2

I got this far, not sure if I'm on the correct path

Yes, that's correct.

Use the following property to simplify it:

a/b ÷ c/d = a/b * d/c

 

[lookagain]

..., but let's remember that L’Hôpital's Rule works with indeterminant rational forms (eg: 0/0, ∞/∞, > > 0/∞, ∞/0 ** < <). Those two are not indeterminate forms.

In this exercise, evaluating the limit directly gives an indeterminant product form (∞*0).

Here is a longer list of indeterminate forms for limits:

\(\displaystyle 0^0, \ \ oo/oo, \ \ 0/0 , \ \ 1^{oo}, \ \ (0)(\pm oo), \ \ oo - oo, \ \ and \ \ oo^0. \)



** Their limits are, respectively, 0 and oo.



Sources:

http://en.wikipedia.org/wiki/Indeterminate_form


http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

 

[Bob Brown MSEE]

Note: d/dx(2atan(1/x)) = -2 /(xx+1)
is -2 when x=0

 

[lookagain]

Note: d/dx > > (2atan(1/x)) < < = -2 /(xx+1)

----> That looks like the equivalent of \(\displaystyle (2a)\tan{\bigg(\dfrac{1}{x}\bigg)}. \ \ \)


is -2 when x=0

.

Alternative suggestions:


d/(dx)[2*arctan(1/x)] = -2/(x^2 + 1)


\(\displaystyle \dfrac{d}{dx}\bigg[2\arctan{\bigg(\dfrac{1}{x}}\bigg)\bigg] \ = \ \dfrac{-2}{x^2 + 1}\)


\(\displaystyle \dfrac{d}{dx}\bigg[2\tan^{-1}{\bigg(\dfrac{1}{x}}\bigg)\bigg] \ = \ \dfrac{-2}{x^2 + 1}\)

 

[Bob Brown MSEE]

Hint

.

Alternative suggestion:
\(\displaystyle \dfrac{d}{dx}\bigg[2\tan^{-1}{\bigg(\dfrac{1}{x}}\bigg)\bigg] \ = \ \dfrac{-2}{x^2 + 1}\)

Let u=1/x and take limit u->0, with L'Hopital's rule and using ...

\(\displaystyle \dfrac{d}{du}\bigg[2\tan^{-1}{\bigg(\dfrac{1}{u}}\bigg)\bigg] \ = \ \dfrac{-2}{u^2 + 1}\)

(is the point of the hint that I posted).