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L’ Hospital’s Rule We have dealt with problems which had indeterminate from either 0/0 or ∞/∞ . The other indeterminate forms are ∞-∞,0,∞,00,∞0,1∞ We state below a rule, called L' Hospital's Rule, meant for problems on limit of the form 0/0 or ∞/∞ . Let f(x) and g(x) be functions differentiable in the neighbourhood of the point a, except may be at the point a itself. If limx→a f(x) = 0 = limx→a g(x) or limx→af(x)= ∞ = ∞ g(x), then limx→a f(x)/g(x) = limx→a f' (x)/g(x) = limx→a f' (x)/g'(x) provided that the limit on the right either exists as a finite number or is ± ∞ . Illustration: Evaluate limx→1 (1-x+lnx)/(1+cos π x ) Solution: limx→1 (1-x+lnx)/(1+cos π x ) (of the form 0/0) = limx→1 (1-1/x)/(-π sin π x) (still of the form 0/0) = limx→1 (x-1)/(πx sin π x) (algebraic simplification) = limx→1 1/(πx sin π x + π2 x cos π x ) (L' Hospital's rule again) = - 1/π2 Illustration: Evaluate limx→y (xy-yx)/(xx-yy ) Solution: limx→y (xy-yx)/(xx-yy ); [0/0] = limx→y (yxy-1 - yx log y)/(xx log(ex) ) = (1-log y)/log(ey) To read more, Buy study materials of Limtis and Continuity comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
We have dealt with problems which had indeterminate from either 0/0 or ∞/∞ .
The other indeterminate forms are ∞-∞,0,∞,00,∞0,1∞
We state below a rule, called L' Hospital's Rule, meant for problems on limit of the form 0/0 or ∞/∞ .
Let f(x) and g(x) be functions differentiable in the neighbourhood of the point a, except may be at the point a itself. If limx→a f(x) = 0 = limx→a g(x) or limx→af(x)= ∞ = ∞ g(x), then limx→a f(x)/g(x) = limx→a f' (x)/g(x) = limx→a f' (x)/g'(x) provided that the limit on the right either exists as a finite number or is ± ∞ .
Illustration:
Evaluate limx→1 (1-x+lnx)/(1+cos π x )
Solution:
limx→1 (1-x+lnx)/(1+cos π x ) (of the form 0/0)
= limx→1 (1-1/x)/(-π sin π x) (still of the form 0/0)
= limx→1 (x-1)/(πx sin π x) (algebraic simplification)
= limx→1 1/(πx sin π x + π2 x cos π x ) (L' Hospital's rule again)
= - 1/π2
Evaluate limx→y (xy-yx)/(xx-yy )
limx→y (xy-yx)/(xx-yy ); [0/0] = limx→y (yxy-1 - yx log y)/(xx log(ex) )
= (1-log y)/log(ey) To read more, Buy study materials of Limtis and Continuity comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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