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) )