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
AJ
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 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#

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