Difference between revisions of "1967 AHSME Problems/Problem 23"
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== See also == | == See also == | ||
| − | {{AHSME box|year=1967|num-b=22|num-a=24}} | + | {{AHSME 40p box|year=1967|num-b=22|num-a=24}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:40, 16 August 2023
Problem
If 
is real and positive and grows beyond all bounds, then 
approaches:
Solution
Since 
, the expression is equal to 
.
The expression 
is equal to 
. As gets large, the second term approaches 
, and thus approaches 
. Thus, the expression approaches 
, which is 
.
Alternately, we divide the numerator and denominator of by to get 
. As grows large, both fractions approach , leaving 
, and so the expression approaches 
.
With either reasoning, the answer is 
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
