Mathematics 116 HWK 14 Solutions Section 4.5 p305. Note: This set of solutions also includes 3 problems from HWK 12 (5,7,11 from 4.5).

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1 Mathematics 6 HWK 4 Solutions Section 4.5 p305 Note: This set of solutions also includes 3 problems from HWK 2 (5,7, from 4.5). Find the indicated it. Use l Hospital s Rule where appropriate. Consider using more elementary methods. If l Hospital s Rule doesn t apply, eplain why. Problem 5, 4.5, p Solution. This it is of the form 0. L Hospital s Rule would apply, and this problem could 0 be done that way, but it s much simpler to use a little algebra. (Another alternative would be to recognize this epression as a derivative, but that s probably more time consuming on this one.) We have 2 + = ( )( + ) = ( ) = 2 + Problem 7, 4.5, p e sin Solution. This it is of the form 0 0, and there s no way to rewrite it, so try l Hospital s Rule. You ll find that one application of l Hospital does the job. e 0 sin = e 0 cos = Problem 9, 4.5, p tan p tan q Solution. Since tan 0 = 0 and the tangent function is continuous wherever it s defined, this it is of the form 0 0. Use l Hospital, as follows: tan p 0 tan q = psec 2 p 0 q sec 2 q = p sec2 0 q sec 2 0 = p q = p q In making this calculation we ve used the continuity of the secant function at 0, as well as the fact that sec 0 =. Page of 7 A. Sontag October 27, 2002

2 Math 6 HWK 4 Solns continued Problem 0, 4.5, p305. Find π tan Solution. Simply use it laws and the continuity of the tangent function at π: tan = 0 π π = 0 Problem, 4.5, p305. ln 0 + Solution. Since ln = and = 0, we can determine the it from the form we already have. Notice, also, that the numerator is negative and the denominator is positive for > 0. Putting this information together, we conclude that ln 0 + = L Hospital s Rule does not apply since we have neither 0 0 nor. In fact, using l Hospital s Rule would lead us to a faulty conclusion. ln ln Problem 2, 4.5, p305. Find Solution. As, ln and so ln(ln )) as well. The given it is therefore of the form, which is indeterminate. Since there s no obvious way to rewrite, use l Hospital s Rule: ln(ln )) ln = = 0 Thus the function y = grows more rapidly than y = ln(ln ), as. Page 2 of 7 A. Sontag October 27, 2002

3 Math 6 HWK 4 Solns continued Problem 6, 4.5, p305. Find 0 cos m cos n 2 Solution. As it stands, this it is of the form 0 0. Apply l Hospital s Rule (twice, after verifying that the first application of l Hospital also yields a it of the form 0 0 : cos m cos n m sin m + nsin n 0 2 = 0 2 m 2 cos m + n 2 cos n = 0 2 m 2 + n 2 = 0 2 Of course, we ve also used it laws and the continuity of the sine and cosine functions. Problem 9, 4.5, p305. ln( + 2e ) Solution. As, we have 2e, so ln( + 2e ). Thus this it is of the form, amenable to l Hospital s Rule. Applying l Hospital and simplifying, we have ln( + 2e ) = + 2e 2e = +2e 2e This new it is also of the form. We can either use a little algebra (factor out 2e from both numerator and denominator) or we can use l Hospital again. Since I m guessing you used l Hospital, I ll do that as well. This gives + 2e 2e 2e = 2e = Now combine these results to get the value of the original it: ln( + 2e ) = + 2e = 2e Page 3 of 7 A. Sontag October 27, 2002

4 Math 6 HWK 4 Solns continued Problem 22, 4.5, p305. Find 2 e. Solution. The given it is indeterminate of the form 0. A little bit of algebra will bring it into the form, and then we can use l Hospital s Rule. As it turns out, we ll need to use l Hospital twice, both times to the form. Here are the details. 2 e = 2 = e 2 = e 2 e = 0 Problem 23, 4.5, p305. e ln Solution. As, e 0 and ln, so this it is of the form 0. A little algebra gives ln e ln = e and the it on the right-hand side is of the form. We can now apply l Hospital to find ln e ln = e = e = 0 since, as, we have 0 and e. Page 4 of 7 A. Sontag October 27, 2002

5 Math 6 HWK 4 Solns continued ( ) Problem 27, 4.5, p305. Find 0 csc Solution. For > 0, values of that are nearly 0 will make both and csc be large positive, hence make the it in question be of the form. Similar reasoning gives the form + for 0. So the given it has the indeterminate form. Rewrite it as a quotient and use l Hospital s Rule. Here s one way to do that (note that I m using l Hospital twice, after verifying that the first application of l Hospital still yields the form 0 0 : ( ) ( 0 csc = 0 ) sin sin = 0 sin cos = 0 sin + cos sin = 0 cos + cos sin = 0 2 = 0 Problem 29, 4.5, p305. e Solution. As, 0, e, and e. Thus this it is of the form, which is indeterminate. As a first step, factor out the : e = (e ) The new it is of the form 0, also indeterminate. Use a little more algebra to bring it into the form 0 0, and then use l Hospital: (e e ) = = (e )( ) 2 = e = e 0 = 2 Page 5 of 7 A. Sontag October 27, 2002

6 Math 6 HWK 4 Solns continued Problem 3, 4.5, p305. Find 0 + sin Solution. As 0 +, sin 0 +, so this it is of the form 0 0. This form is indeterminate, as the zero in the base tries to push the whole epression to 0, while the zero in the eponent tries to push the whole epression to. This is eactly the sort of it where we use l Hospital s Rule in combination with the logarithm function and then e it up. In this case, we ll need to do a little bit of algebra to bring ln( sin ) into the form of a quotient, and the quotient we get will be of the form : () ln(sin ln ) = sin ln = csc = 0 + csc cot This last it is still of the form. First rewrite it using trigonometric identities, and then we ll analyze further. 0 + csc cot = sin tan 0 + sin If you remember the special it =, then you can see that we ll have 0 sin tan (2) = (sin ) ) = 0 = 0 0 +( tan Otherwise, use l Hospital again to find (2 sin tan sin sec 2 cos tan ) = = Combine () with either (2) or (2 ) to obtain ln(sin ) = and e it up to find sin sin = ) eln( = e ln(sin ) 0 + = e 0 = Page 6 of 7 A. Sontag October 27, 2002

7 Math 6 HWK 4 Solns continued Problem 33, 4.5, p305. Find 0 ( 2) Solution. As 0, ±, so this it is of the form ±, which is indeterminate. In fact, when is just a bit above 0, we ll have a number just less than raised to a huge power, and there ll be a tug of war between an overall value of (because the base is nearly ) and 0 (because a number smaller than is being raised to a huge power). When is just a bit below 0, we ll have a number just above raised to a huge negative power, which also, in the end, gives a tug of war between and 0. Once again we use l Hospital s Rule in combination with the logarithm function and then e it up. In this case, we ll be using l Hospital on a quotient that is of the form 0 0. Here are the details. Now e it up to find that 0 ln( 2) 0 ( 2) ln( 2) = = = 2 = e ln(( 2) ) = e ln( 2) 0 = e 2. 0 Page 7 of 7 A. Sontag October 27, 2002