1967 AHSME Problems/Problem 23
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 | |
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