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Give the proof of LH rule when f(a)=0 and g(a)=0
f(x) is the numerator and g(x) is the denominator...
Trust me it is not very difficult..
Use the basic definition of Derivative...
(Sorry mistake galti ho gaya.)
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16 Answers
24but sir why would we apply LH rule then?????????????????because it wont be indeterminate then.....
1if both are in num. then the limit should be 0 :) what was to prove...........?........?....?
62sorry galti ho gaya...
bhawna ko samjo ;)
aise haan this was not the proof i was looking for eureka.. but still this is good enuf...
There is a more "elegant" proof...
33If we take derivative of both num and den... we cancel 0 making factor.. if both have only one root at that pt... diff 1nce is enough if two repeated roots then f'(x) will have one root so f'(a) will be also 0 so we have to diff again ...
11Ltx→af(x)/g(x) =Lt h→0 f(a+h)/g(a+h)
=Lth→0 f(a+h)-f(a)/g(a+h)-g(a) ((using g(a)=f(a)=0))
=Lth→0 (f(a+h)-f(a)/h)/(g(a+h)-g(a)/h)
=Lt x→a f'(x)/g'(x)
11was this what u where expecting or r u expecting some thing other than this
1bhaiya aap aise 'also' likhke bhawnao ko zaahir kiye ki hum sab samajh ke bhi soch rahe the ... ki jo hum samjhe woh hume nahi samajhna chaiye tha.. [3] [3]
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