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  lim_x→a f(x) = 0 = lim_x→a g(x) or lim_x→af(x)= ∞ = ∞  g(x), then lim_x→a f(x)/g(x) = lim_x→a f' (x)/g(x) = lim_x→a f' (x)/g'(x)    provided that the limit on the right either exists as a finite number or is ± ∞ .

Illustration:

Evaluate  lim_x→1 (1-x+lnx)/(1+cos π x )

Solution:

  lim_x→1 (1-x+lnx)/(1+cos π x ) (of the form 0/0)

= lim_x→1 (1-1/x)/(-π sin π  x) (still of the form 0/0)

=  lim_x→1 (x-1)/(πx sin π x) (algebraic simplification)

=  lim_x→1 1/(πx sin π x + π² x cos π x ) (L' Hospital's rule again)

 = - 1/π²

Illustration:     

Evaluate lim_x→y  (x^y-y^x)/(x^x-y^y )

Solution:            

 lim_x→y  (x^y-y^x)/(x^x-y^y );   [0/0] = lim_x→y (yx^y-1 - y^x log y)/(x^x log(ex) )

                                                     = (1-log y)/log(ey)
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