±. Then. . x. lim g( x) = lim. cos x 1 sin x. and (ii) lim

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1 MATH 36 L'H ˆ o pital s Rule Si of the indeterminate forms of its may be algebraically determined using L H ˆ o pital's Rule. This rule is only stated for the / and ± /± indeterminate forms, but four other indeterminate forms may be manipulated into one of these two forms. Theorem. (L H o ˆ pital's Rule) Let f ( ) and g( ) be differentiable functions such that f ( ) g( ) yields an indeterminate form of or ± ±. Then a a f ( ) g( ) = a f ʹ ( ) g ʹ ( ). sin Eample. Evaluate (i) and (ii) cos. Solution. (i) By substituting =, we obtain sin =. So we separately take the derivatives of the numerator and the denominator, and then reevaluate the it: sin d(sin ) / d = d( ) / d (ii) Likewise, for = we obtain cos cos sin have = = sin =. = cos = = =. =. By L H o ˆ pital's Rule, we then Eample. Evaluate ln. Solution. Both of the functions ln and grow very slowly to + as. So ln initially we have, =. We now separately take the derivatives of the numerator and the denominator, and we simplify the result before reevaluating the it. ln = d(ln ) / d d( ) / d = = = = =. So even though ln and both grow to + as, the function grows more quickly and therefore forces the fraction ln to.

2 Eample 3. Evaluate e. Solution. Both of the functions e and grow very quickly to + as. So e initially we have, =. We now separately take the derivatives of the numerator and the denominator, and reevaluate the it. e = d (e ) / d d( ) / d = e 9 =. We still obtain the / form. So we re-apply L H ˆ o pital's Rule over and over until we obtain a determined form: e = e 9 = e 9 8 =... = e! =! = + So even though e and both grow to + as, the function e grows more quickly and therefore forces the fraction e to +. From Eample 3, we can see the following result: Theorem. Let c > and n >. Then (a) e c n = + and (b) n e c = n c =. e In other words, eponential growth dominates polynomial growth. The ± Form The ± indeterminate form can be converted to / or ± /± by dividing by the reciprocal of one of the functions: a f ( ) g( ) = a g( ) f ( ) Thus we may obtain ± = ± = ± ± or ± = ± =.

3 Eample 4. Evaluate the it sin( ) tan. Solution. By substituting = π /, we obtain sin(π) tan(π / ) = +. We now have two choices as to how to re-write the function: sin( )tan = tan sin( ) = tan ±, which will yield a csc( ) ± form, or sin( )tan = sin( ) tan = sin( ) cot, which will yield a form. We shall apply L H ˆ o pital's Rule to both forms. In both cases, we take the derivatives of the numerator and the denominator, then simplify the result before trying to re-evaluate the it. sin( ) tan = tan csc( ) = d(tan ) / d d(csc( )) / d = sec csc( ) cot( ) = sin ( ) cos cos( ) = ( sin cos ) cos cos( ) = sin cos( ) = sin (π / ) cos( π) = =. In the second case, the simplification is a little easier and we have sin( ) tan = sin( ) cot = d(sin( )) / d d(cot ) / d = cos( ) csc = cos( ) π sin = cos(π) sin (π / ) =. Eample 5. Evaluate + n ln.

4 Solution. Initially, as +, we obtain the form. In this case we re-write the function leaving ln in the numerator, which will yield a / form. Then we apply L H o ˆ pital's Rule to obtain a it of : + n ln = ln + n = = + n n+ = d (ln ) / d + d( n ) / d n+ + n = + n n =. The Eponential Indeterminate Forms The indeterminate forms,, and can be converted to the ± form by taking the natural logarithm: ln( ) = ln = ln( ) = ln = ln( ) = ln = We then convert these forms to either / or ± /± by dividing by the reciprocal of one of the functions. Generally it is best to keep the logarithm in the numerator and divide by the reciprocal of the other function. Then we apply L H o ˆ pital's Rule to obtain a it L. Finally, we undo the natural logarithm and say that e L is the it of the original function. Eample 6. Evaluate (tan )cos. Solution. Initially we obtain the form. Taking the natural logarithm gives us the form ln( ) = ln = ln / =. Applying this result to the functions, we have ln(tan ) cos ln(tan ) = cos ln(tan ) = / cos = ln(tan ). We now apply L H o sec ˆ pital's Rule by taking the derivatives of the numerator and the denominator of this last epression: ln(tan ) sec = sec tan sec tan = sec tan = cos sin = =. Finally, we undo the logarithm with this it of to obtain (tan )cos = e =.

5 Eample 7. Evaluate (cos )/. + Solution. Initially we obtain the form. Taking the natural logarithm gives us the form ln( ) = ln = = / =. Applying this result to the functions, we have ln(cos ) / = ln(cos ) ln(cos ) =. We now apply L H o ˆ pital's Rule by taking the derivatives of the numerator and the denominator of this last epression: sin ln (cos ) + = cos = + sin + cos = =. Finally, we undo the logarithm with this it of / to obtain (cos )/ = e /. + Eample 8. Evaluate (sin(π ))ln. Solution. Initially we obtain the form. Taking the natural logarithm gives us the form ln( ) = ln = = / =. Applying this result to the functions, we have ln(sin(π )) ln ln(sin(π )) ln (sin(π )) = ln ln(sin(π )) = = (ln ). We now apply ln L H o ˆ pital's Rule by taking the derivatives of the numerator and the denominator of this last epression: ln(sin(π )) (ln ) = π cos(π ) sin(π ) (ln ) π cos(π )(ln ) = sin(π ) (ln ) = π cos(π ) = π sin(π ) (ln ) sin(π ) = π. We now apply L H ˆ o pital's Rule again: (ln ) π sin(π ) = π (ln )( / ) (ln ) = π cos( π ) cos(π ) = ln(sin(π )) (ln ) Finally, we have (sin(π ))ln = e =. = π (ln ) sin(π ) =. = ; thus,

6 We conclude with one of the most famous results in calculus. Theorem. For all real numbers a, Proof. As, we have + a + a = + a = (+ ) =, which is an indeterminate form. We now take the natural logarithm, which gives the form as follows: Net apply L H ˆ o pital's Rule to denominator: ln + a = ln + a ln + a = ln + a + a a = a + a as ln =. by taking derivatives of numerator and = a. as Finally, we undo the logarithm with this it to obtain + a For eample, + 5 = e 5 and 4 = e /4. Corollary. The number e is given by + = e. Corollary. For all real numbers a, + a Proof. We simply have + a = + a