MHB Prove L'Hopitals rule using Taylors expansion

[ognik]

I'm given that $ \frac{f\left({x}_{o}\right)}{g\left({x}_{o}\right)} =\frac{0}{0} $ and asked to show that:

$ \lim_{{x}\to{{x}_{0}}}\frac{f\left({x}\right)}{g\left({x}\right)} = \lim_{{x}\to{{x}_{0}}} \frac{f^{'} \left({x}\right)}{g^{'}\left({x}\right)} $
This should be easy, but I suspect I've not got it quite right. Using Taylor (not Maclauren):

$ f\left(x\right) =f\left({x}_{0}\right) + {f}^{'}\left({x}_{0}\right)\left(x - {x}_{0}\right) + {f}^{''}\frac{\left(x-{x}_{0}\right)^2}{2!} + ... $

1) Am I correct to be able to ignore terms in $ x^2 $ and higher powers - because $ \left(x-{x}_{0}\right) $ is very small in the limit?

Then I get $ \lim_{{x}\to{{x}_{0}}} \frac{f\left({x}\right)}{g\left({x}\right)} =
\lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}_{0}\right)\left(x-{x}_{0}\right)} {{g}^{'}\left({x}_{0}\right)\left(x-{x}_{0}\right)} =
\lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}_{0}\right)} {{g}^{'}\left({x}_{0}\right)} $

2) Can I argue, by 'reversing' $ _{{x}\to{{x}_{0}}} $, to let $ \lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}_{0}\right)} {{g}^{'}\left({x}_{0}\right)} =
\lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}\right)} {{g}^{'}\left({x}\right)} $

I get the feeling what I've done is clumsy and there is a better way?

Thanks for reading

 

[Sudharaka]

I'm given that $ \frac{f\left({x}_{o}\right)}{g\left({x}_{o}\right)} =\frac{0}{0} $ and asked to show that:

$ \lim_{{x}\to{{x}_{0}}}\frac{f\left({x}\right)}{g\left({x}\right)} = \lim_{{x}\to{{x}_{0}}} \frac{f^{'} \left({x}\right)}{g^{'}\left({x}\right)} $
This should be easy, but I suspect I've not got it quite right. Using Taylor (not Maclauren):

$ f\left(x\right) =f\left({x}_{0}\right) + {f}^{'}\left({x}_{0}\right)\left(x - {x}_{0}\right) + {f}^{''}\frac{\left(x-{x}_{0}\right)^2}{2!} + ... $

1) Am I correct to be able to ignore terms in $ x^2 $ and higher powers - because $ \left(x-{x}_{0}\right) $ is very small in the limit?

Then I get $ \lim_{{x}\to{{x}_{0}}} \frac{f\left({x}\right)}{g\left({x}\right)} =
\lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}_{0}\right)\left(x-{x}_{0}\right)} {{g}^{'}\left({x}_{0}\right)\left(x-{x}_{0}\right)} =
\lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}_{0}\right)} {{g}^{'}\left({x}_{0}\right)} $

2) Can I argue, by 'reversing' $ _{{x}\to{{x}_{0}}} $, to let $ \lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}_{0}\right)} {{g}^{'}\left({x}_{0}\right)} =
\lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}\right)} {{g}^{'}\left({x}\right)} $

I get the feeling what I've done is clumsy and there is a better way?

Thanks for reading

Yes that's correct. Alternatively you can factor the $x-x_0$ part and cancel it out.

\[\lim_{{x}\to{{x}_{0}}} \frac{f\left({x}\right)}{g\left({x}\right)} =
\lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}_{0}\right)\left(x-{x}_{0}\right)+\frac{f''(x_0)(x-x_0)^2}{2!}+\cdots} {{g}^{'}\left({x}_{0}\right)\left(x-{x}_{0}\right)+\frac{g''(x_0)(x-x_0)^2}{2!}+\cdots} = \lim_{{x}\to{{x}_{0}}} \frac{{f}^{'}\left({x}_{0}\right)+\frac{f''(x_0)(x-x_0)}{2!}+\cdots} {{g}^{'}\left({x}_{0}\right)+\frac{g''(x_0)(x-x_0)}{2!}+\cdots} = \frac{{f}^{'}\left({x}_{0}\right)} {{g}^{'}\left({x}_{0}\right)}\]

If $f'$ and $g'$ are continuous (in other words $f$ and $g$ are continuously differentiable) we have,

\[\lim_{{x}\to{{x}_{0}}} \frac{f\left({x}\right)}{g\left({x}\right)} = \lim_{x\rightarrow x_0}\frac{f'(x)}{g'(x)}\]

Hope this helps. :)

 

[ognik]

Thanks :-)

 

[Sudharaka]

Thanks :-)

You are welcome. :)