Mixing
How does $dfrac{f(x)}{g(x)}approx dfrac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)} implies dfrac{f'(a)}{g'(a)}$?
$begingroup$
Can someone please unveil the steps for this answer?
Thus
$$
frac{f(x)}{g(x)}approx color{red}{frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}}.
$$
Taking the limit of the right hand side gives $dfrac{f'(a)}{g'(a)}$.
Because $x to a iff x- a to 0$, then
$$lim_{x to a} color{red}{dfrac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}} = dfrac{f'(a)times0 + f(a)}{g'(a)times0 + g(a)}$$
limits
edited Jan 12 at 3:20
Eevee Trainer
5,9571936
asked Jan 12 at 3:18
Tucker RapuTucker Rapu
61211037
$endgroup$
add a comment |
$begingroup$
Can someone please unveil the steps for this answer?
Thus
$$
frac{f(x)}{g(x)}approx color{red}{frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}}.
$$
Taking the limit of the right hand side gives $dfrac{f'(a)}{g'(a)}$.
Because $x to a iff x- a to 0$, then
$$lim_{x to a} color{red}{dfrac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}} = dfrac{f'(a)times0 + f(a)}{g'(a)times0 + g(a)}$$
limits
edited Jan 12 at 3:20
Eevee Trainer
5,9571936
asked Jan 12 at 3:18
Tucker RapuTucker Rapu
61211037
$endgroup$
5
$begingroup$
In that reference, one is applying this to the case $f(a)=f(b)=0$ (in order to justify the Hospital).
$endgroup$
– Lord Shark the Unknown
Jan 12 at 3:29
add a comment |
$begingroup$
Can someone please unveil the steps for this answer?
Thus
$$
frac{f(x)}{g(x)}approx color{red}{frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}}.
$$
Taking the limit of the right hand side gives $dfrac{f'(a)}{g'(a)}$.
Because $x to a iff x- a to 0$, then
$$lim_{x to a} color{red}{dfrac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}} = dfrac{f'(a)times0 + f(a)}{g'(a)times0 + g(a)}$$
limits
edited Jan 12 at 3:20
Eevee Trainer
5,9571936
asked Jan 12 at 3:18
Tucker RapuTucker Rapu
61211037
$endgroup$
Can someone please unveil the steps for this answer?
Thus
$$
frac{f(x)}{g(x)}approx color{red}{frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}}.
$$
Taking the limit of the right hand side gives $dfrac{f'(a)}{g'(a)}$.
Because $x to a iff x- a to 0$, then
$$lim_{x to a} color{red}{dfrac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}} = dfrac{f'(a)times0 + f(a)}{g'(a)times0 + g(a)}$$
limits
limits
edited Jan 12 at 3:20
Eevee Trainer
5,9571936
asked Jan 12 at 3:18
Tucker RapuTucker Rapu
61211037
edited Jan 12 at 3:20
Eevee Trainer
5,9571936
asked Jan 12 at 3:18
Tucker RapuTucker Rapu
61211037
edited Jan 12 at 3:20
Eevee Trainer
5,9571936
edited Jan 12 at 3:20
Eevee Trainer
5,9571936
edited Jan 12 at 3:20
Eevee Trainer
5,9571936
5,9571936
asked Jan 12 at 3:18
Tucker RapuTucker Rapu
61211037
asked Jan 12 at 3:18
Tucker RapuTucker Rapu
61211037
asked Jan 12 at 3:18
Tucker RapuTucker Rapu
61211037
61211037
5
$begingroup$
In that reference, one is applying this to the case $f(a)=f(b)=0$ (in order to justify the Hospital).
$endgroup$
– Lord Shark the Unknown
Jan 12 at 3:29
add a comment |
5
$begingroup$
In that reference, one is applying this to the case $f(a)=f(b)=0$ (in order to justify the Hospital).
$endgroup$
– Lord Shark the Unknown
Jan 12 at 3:29
5
5
$begingroup$
In that reference, one is applying this to the case $f(a)=f(b)=0$ (in order to justify the Hospital).
$endgroup$
– Lord Shark the Unknown
Jan 12 at 3:29
$begingroup$
In that reference, one is applying this to the case $f(a)=f(b)=0$ (in order to justify the Hospital).
$endgroup$
– Lord Shark the Unknown
Jan 12 at 3:29
add a comment |
0
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5
$begingroup$
In that reference, one is applying this to the case $f(a)=f(b)=0$ (in order to justify the Hospital).
$endgroup$
– Lord Shark the Unknown
Jan 12 at 3:29