Help me to proof this L'Hospital's Rule II for case (b) please? I am waiting for your help. Thank you very much for your help before.

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1 Answer
Apr 23, 2018

The proofs of both cases are mentioned below, one for 0/0 and oo/oo

Explanation:

For 0/0
Let a function h (x)=(f (x))/(g (x))

To find limit at xrarra. L'Hôpitals' rule only works when directly substituted in f (x) and g (x) results in 0.
Hence assume, f (a)=g (a)=0.
Hence,
=Lim_(xrarra) f (x)/g (x)
=Lim_(xrarra) (f(x)-0)/(g (x)-0)
=Lim_(xrarra) [f (x)-f (a)]/[g (x)-g (a)] .:f (a)=g (a)=0
Multiply by (1/(x-a))/(1/(x-a))
=Lim_(xrarr) [f (x)-f (a)]/[g (x)-g (a)] × (1/(x-a))/(1/(x-a))
=Lim_(xrarra) [f (x)-f (a)]/(x-a)/[g (x)-g (a)]/(x-a)
Use division law of limit,
=[Lim_(xrarra) [f (x)-f (a)]/(x-a)]/[Lim_(xrarra) [g (x)-g (a)]/(x-a)]

It is the definition of derivatives at x=a Hence,
=(f'(a))/(g'(a))
:.Lim_(xrarra) h (x)=f (x)/g (x)=(f'(a))/(g'(a))

For second case as you asked It follows the same proof just by a slight modification and the rest proof is just replica of above proof,
To proove foroo/oo
We know a/b =(1/b)/(1/a)
So,
0/0 =(1/0)/(1/0)
.:1/0=oo
:.0/0=oo/oo
Hence we can show the same proof for oo/oo as for 0/0