= , then prove that m = ± . As this is an indeterminate form , using the L’Hôpital’s Rule lim θ θ θ = lim θ θ θ Now using the example . , we have lim θ θ θ θ θ × = m Therefore m = n That is m = ± n . Example .
Evaluate : lim log( cot( → − This is an indeterminate form ¥ ¥ and hence we use the L’Hôpital’s Rule to evaluate the limit. Thus, lim log( cot( → − = lim ∞ ∞ cosec form On simplication, we get = lim sin ( → − . form again applying the L’Hôpital’s Rule, we get = lim sin( ) cos( → − ⋅ π = lim sin( ) cos( → − − ⋅ = . Example .
Evaluate : lim → + . This is an indeterminate of the form ∞−∞ . To evaluate this limit we first simplify and bring it in the form and applying the L’Hôpital’s Rule, we get lim → + = lim ) . x e → + form Now, lim x e → + = lim xe → + form = lim xe → + = .
Applications of Differential Calculus Example . Evaluate : lim → + This is an indeterminate of the form ( ×∞ . To evaluate this limit, we first simplify and bring it to the form ∞ ∞ and apply L’Hôpital’s Rule. Thus, we get lim → + = lim → + ∞ ∞ form = lim → + − = lim( → + − .
Example . Evaluate : lim →∞ . This is an indeterminate of the form ∞ ∞ . To evaluate