Question

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.

Answer 1

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 for`oo/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`