Mathematics Asked by BlackThunder on November 24, 2021
I was trying to prove L’Hospital’s rule when $L= displaystyle limlimits_{x to a}dfrac {f(x)}{g(x)}$=$dfrac {infty}
{infty}$.
So this is what I currently tried to prove that this $L= displaystyle limlimits_{x to a} dfrac {f(x)}{g(x)}$ is equal to $displaystyle lim_{xto a}dfrac {f'(x)}{g'(x)}$ .
This is what I have tried, $displaystyle lim_{x to a} dfrac {1/g(x)}{1/f(x)}=frac {0}{0}$.
Now I know this can be proved by the baby version, but I would like to see if I can do it without that.
So when we know that $dfrac{1}{g(a)}=0$ and $dfrac{1}{f(a)}=0$. So ${}=displaystyle limlimits_{x to a}dfrac {1/g(x)-1/g(a)}{1/f(x)-1/f(a)}$.
I just do not know what to do after this, I tried taking the LCM of $g(x)$ and $g(a)$ but I could not get anywhere. If anyone could help me proving this without using the baby method and rather use the meaning of a differential which is what proves the baby method, $${}= displaystyle limlimits_{x to a} dfrac {frac {f(x)-f(a)}{x-a}}{frac {g(x)-g(a)}{x-a}}=dfrac {f'(x)}{g'(x)}$$. Any help would be really appreciated.
Hint:
For any $varepsilon>0$, there exists $a_varepsilon>0$ such that $Big|frac{f'(x)}{g'(x)}-LBig|<varepsilon$ whenever $a<xleq a_varepsilon$.
For any $a<x<y<a_varepsilon$, an application of the mean value theorem gives
$$ begin{align} frac{f(y)-f(x)}{g(y)-g(x)}&=frac{f'(c_{x,y})}{g'(c_{x,y})} end{align} $$ for some $x<c_{x,y}< y$. From $$ begin{align} frac{f(y)-f(x)}{g(y)-g(x)}&=frac{f(x)}{g(x)}frac{1-frac{f(y)}{f(x)}}{1-frac{g(y)}{g(x)}} end{align} $$ we obtain $$ begin{align} frac{f(x)}{g(x)}= frac{f'(c_{x,y})}{g'(c_{x,y})}frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}} end{align} $$
$$ begin{align} frac{f(x)}{g(x)}&= Big(frac{f'(c_{x,y})}{g'(c_{x,y})} -L +LBig)left(frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}}-1+1right)\ &=Big(frac{f'(c_{x,y})}{g'(c_{x,y})} -LBig)left(frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}}-1right) + Lleft(frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}}-1right) + \& quadquad Big(frac{f'(c_{x,y})}{g'(c_{x,y})} -LBig) + L end{align} $$
and so
$$ begin{align} frac{f(x)}{g(x)} - L&= Big(frac{f'(c_{x,y})}{g'(c_{x,y})} -L +LBig)left(frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}}-1+1right) - L\ &=Big(frac{f'(c_{x,y})}{g'(c_{x,y})} -LBig)left(frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}}-1right) + Lleft(frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}}-1right) + Big(frac{f'(c_{x,y})}{g'(c_{x,y})} -LBig) end{align} $$
The factor $Big|frac{f'(c_{x,y})}{g'(c_{x,y})} -LBig|<varepsilon$ since $a<x<c_{x,y}<y<a_varepsilon$. By letting $xrightarrow a$, we get $$ left(frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}}-1right)xrightarrow{xrightarrow a}0 $$ and so, $$ limsup_{xrightarrow a}Big|frac{f(x)}{g(x)}-LBig|leq 2varepsilon $$ for all $varepsilon>0$. This implies that $lim_{xrightarrow a}frac{f(x)}{g(x)}$ exists and equals $L$.
If you are not comfortable with $limsup$'s then you can try to further exploit $lim_{xrightarrow a}f(x)=infty=lim_{xrightarrow a}g(x)$ the show that the factor
$$ left(frac{1-frac{g(y)}{g(x)}}{1-frac{f(y)}{f(x)}}-1right) $$
is less than some small term depending on $varepsilon$.
Answered by Oliver Diaz on November 24, 2021
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