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 l’Hôpital’s Rule Help
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Indeterminate Forms Practice Test

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.

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.

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.

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

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

(a) lim _{x → 0} (cos x − 1) = 0 and lim _{x → 0} x ^{2} − x ^{3} = 0 so l’Hôpital’s Rule applies. Thus
Now l’Hôpital’s Rule applies again to yield
(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
l’Hôpital’s Rule applies again to yield
(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
(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
Now l’Hôpital’s Rule applies again to yield
We apply l’Hôpital’s Rule one last time to obtain
(f) lim _{x →1} ( e ^{x} − 1) = 0 and lim _{x →1} ( x − 1) = 0 so l’Hôpital’s Rule applies. Thus

(a) lim _{x →1 + ∞} x ^{3} = lim _{x → + ∞} ( e ^{x} − x ^{2} ) = + ∞ so l’Hôpital’s Rule applies. Thus
l’Hôpital’s Rule applies again to yield
l’Hôpital’s Rule applies one more time to finally yield
(b) lim _{x → + ∞} ln x = lim _{x → + ∞} x = + ∞ so l’Hôpital’s Rule applies. Thus
(c) lim _{x → + ∞} e ^{− x} = lim _{x → + ∞} ln[ x / ( x + 1)] = 0 so l’Hôpital’s Rule applies. Thus
It is convenient to rewrite this expression as
Now l’Hôpital’s Rule applies once more to yield
We apply l’Hôpital’s Rule a last time to obtain
(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
Since lim _{x → − ∞} x = lim _{x → − ∞} e ^{− x} = ±∞, l’Hôpital’s Rule applies. Thus
(f) Since lim _{x → − ∞} ln  x  = lim _{x → −∞} e ^{− x} = + ∞, l’Hôpital’s Rule applies. Thus

(a) We write the limit as . Since lim _{x → + ∞} x ^{3} = lim _{x → + ∞} e ^{x} = + ∞, l’Hôpital’s Rule applies. Thus
We apply l’Hôpital’s Rule again to obtain
Applying l’Hôpital’s Rule one last time yields
(b) We write the limit as . Since lim _{x → + ∞} sin(1/ x ) = lim _{x → + ∞} 1/ x = 0, l’Hôpital’s Rule applies. Hence
(c) We rewrite the limit as . Since lim _{x → + ∞} ln[ x /( x + 1)] = lim _{x → + ∞} 1/( x + 1) = 0, l’Hôpital’s Rule applies. Thus
Now l’Hôpital’s Rule applies again and we obtain
(d) We rewrite the limit as . Since lim _{x → + ∞} ln x = lim _{x → + ∞} e ^{x} = + ∞, l’Hôpital’s Rule applies. Thus
(e) We write the limit as . Since lim _{x → − ∞} lim x ^{2} = lim _{x → −∞} e ^{−2 x} = 0, l’Hôpital’s Rule applies. Thus
l’Hôpital’s Rule applies one more time to yield
(f) We rewrite the limit as . Since lim _{x → 0} e ^{1/ x} = lim _{x →0} 1/ x = +∞, l’Hôpital’s Rule applies. Thus

4 . We do (a), (b), (c), (d).
Therefore
Now this equals
The second limit does not exist, so the original integral does not converage.

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