. . Indeterminate forms , ∞ and ∞ In order to evaluate the indeterminate forms like this, we shall first state the theorem on the limit of a composite function. Theorem .
Let lim ( ) g x →α exist and let it be L and let f x ( ) be a continuous function at x L . Then, lim ( ( )) f g x →α = f g x x lim ( ) α The evaluation procedure for evaluating the limits ( ) Let A g x lim ( ) . Then taking logarithm, with the assumption that A > to ensure the continuity of the logarithm function, we get log limlog( ( )) A g x . Therefore using the above theorem with f x we have the limit limlog( ( )) g x = log lim ( ) a g x ) .
( ) We have the limit limlog( ( )) g x into either or ∞ ∞ form evaluate it using L’Hôpital’s Rule. ( ) Let that evaluated limit be say α . Then the required limit is e α . Example .
Using the L’Hôpital’s Rule, prove that lim( → + This is an indeterminate of the form ¥ . Let g x x x . Taking the logarithm, we get log ( ) g x = log( + x lim log( ( )) g x → + = lim log( → + form = lim (by L’Hôpital’s Rule) = . But, lim log ( ) g x → + = log lim g x → + Therefore, log lim g x → + = .
Hence by exponentiating, we get lim g x → + Example . Evaluate : lim( ) log →∞ This is an indeterminate of the form ¥ . Let g x ( ) = ( ) log + x x . Taking the logarithm, we get log ( ) g x = log( + x Applications of Differential Calculus limlog ( ) g x →∞ = lim log( →∞ ∞ ∞ form = lim →∞ (by L’Hôpital’s Rule) = lim →∞ ∞ ∞ form = →∞ lim but, limlog ( ) g x →∞ = log lim ( ) g x →∞ ) .
Hence by exponentiating, we get the required limit as e . Example . Evaluate : lim Let g x ( ) = . This is an indeterminate of the form ¥ when x → .
Taking the logarithm, log ( ) g x = log x − Therefore, limlog ( ) g x → = lim log form . An application of L’Hôpital’s rule, gives lim = − . But, limlog ( ) g x → = log lim ( ) g x Hence on exponentiating, we get lim = e − = .