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Indeterminate Forms Practice Test

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Review the following concepts if needed:

Indeterminate Forms Practice Test

  1. If possible, use l’Hôpital’s Rule to evaluate each of the following limits. In each case, check carefully that the hypotheses of l’Hôpital’s Rule apply.

    Indeterminate Forms Exercises

  2. If possible, use l’Hôpital’s Rule to evaluate each of the following limits. In each case, check carefully that the hypotheses of l’Hôpital’s Rule apply.

    Indeterminate Forms Exercises

  3. If possible, use some algebraic manipulations, plus l’Hôpital’s Rule, to evaluate each of the following limits. In each case, check carefully that the hypotheses of l’Hôpital’s Rule apply.

    Indeterminate Forms Exercises

    Indeterminate Forms Exercises

  4. Evaluate each of the following improper integrals. In each case, be sure to write the integral as an appropriate limit.

    Indeterminate Forms Exercises

  5. Evaluate each of the following improper integrals. In each case, be sure to write the integral as an appropriate limit.

    Indeterminate Forms Exercises

Answers

  1. (a) lim x → 0 (cos x − 1) = 0 and lim x → 0 x 2x 3 = 0 so l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    Now l’Hôpital’s Rule applies again to yield

    SOLUTIONS TO EXERCISES Chapter 5

    (b) lim x →0 e 2 x − 1 − 2 x = 0 and lim x → 0 x 2 + x 4 = 0 so l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    l’Hôpital’s Rule applies again to yield

    SOLUTIONS TO EXERCISES Chapter 5

    (c) lim x →0 cos x ≠ 0, so l’Hôpital’s Rule does not apply. In fact the limit does not exist.

    (d) lim x →1 [ln x ] 2 = 0 and lim x → 1 ( x − 1) = 0 so l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    (e) lim x →2 ( x − 2) 3 = 0 and lim x →2 sin( x − 2) − ( x − 2) = 0 so l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    Now l’Hôpital’s Rule applies again to yield

    SOLUTIONS TO EXERCISES Chapter 5

    We apply l’Hôpital’s Rule one last time to obtain

    SOLUTIONS TO EXERCISES Chapter 5

    (f) lim x →1 ( e x − 1) = 0 and lim x →1 ( x − 1) = 0 so l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

  2. (a) lim x →1 + ∞ x 3 = lim x → + ∞ ( e xx 2 ) = + ∞ so l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    l’Hôpital’s Rule applies again to yield

    SOLUTIONS TO EXERCISES Chapter 5

    l’Hôpital’s Rule applies one more time to finally yield

    SOLUTIONS TO EXERCISES Chapter 5

    (b) lim x → + ∞ ln x = lim x → + ∞ x = + ∞ so l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    (c) lim x → + ∞ e x = lim x → + ∞ ln[ x / ( x + 1)] = 0 so l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    It is convenient to rewrite this expression as

    SOLUTIONS TO EXERCISES Chapter 5

    Now l’Hôpital’s Rule applies once more to yield

    SOLUTIONS TO EXERCISES Chapter 5

    We apply l’Hôpital’s Rule a last time to obtain

    SOLUTIONS TO EXERCISES Chapter 5

    (d) Since lim x → + ∞ sin x does not exist, l’Hôpital’s Rule does not apply. In fact the requested limit does not exist.

    (e) It is convenient to rewrite this limit as

    SOLUTIONS TO EXERCISES Chapter 5

    Since lim x → − ∞ x = lim x → − ∞ e x = ±∞, l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    (f) Since lim x → − ∞ ln | x | = lim x → −∞ e x = + ∞, l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

  3. (a) We write the limit as SOLUTIONS TO EXERCISES Chapter 5 . Since lim x → + ∞ x 3 = lim x → + ∞ e x = + ∞, l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    We apply l’Hôpital’s Rule again to obtain

    SOLUTIONS TO EXERCISES Chapter 5

    Applying l’Hôpital’s Rule one last time yields

    SOLUTIONS TO EXERCISES Chapter 5

    (b) We write the limit as SOLUTIONS TO EXERCISES Chapter 5 . Since lim x → + ∞ sin(1/ x ) = lim x → + ∞ 1/ x = 0, l’Hôpital’s Rule applies. Hence

    SOLUTIONS TO EXERCISES Chapter 5

    (c) We rewrite the limit as SOLUTIONS TO EXERCISES Chapter 5 . Since lim x → + ∞ ln[ x /( x + 1)] = lim x → + ∞ 1/( x + 1) = 0, l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

     

    Now l’Hôpital’s Rule applies again and we obtain

    SOLUTIONS TO EXERCISES Chapter 5

    (d) We rewrite the limit as SOLUTIONS TO EXERCISES Chapter 5 . Since lim x → + ∞ ln x = lim x → + ∞ e x = + ∞, l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    (e) We write the limit as SOLUTIONS TO EXERCISES Chapter 5 . Since lim x → − ∞ lim x 2 = lim x → −∞ e −2 x = 0, l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

    l’Hôpital’s Rule applies one more time to yield

    SOLUTIONS TO EXERCISES Chapter 5

    (f) We rewrite the limit as SOLUTIONS TO EXERCISES Chapter 5 . Since lim x → 0 e 1/ x = lim x →0 1/ x = +∞, l’Hôpital’s Rule applies. Thus

    SOLUTIONS TO EXERCISES Chapter 5

  4. 4 . We do (a), (b), (c), (d).

    SOLUTIONS TO EXERCISES Chapter 5

    SOLUTIONS TO EXERCISES Chapter 5

     

    SOLUTIONS TO EXERCISES Chapter 5

    Therefore

    SOLUTIONS TO EXERCISES Chapter 5

     

    Now this equals

    SOLUTIONS TO EXERCISES Chapter 5

    The second limit does not exist, so the original integral does not converage.

  5. We do (a), (b), (c), (d).

    SOLUTIONS TO EXERCISES Chapter 5

    SOLUTIONS TO EXERCISES Chapter 5

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