Difference between revisions of "1950 AHSME Problems/Problem 40"
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| − | == | + | ==Solution 2== |
| + | The numerator of <math> \frac {x^2-1}{x-1}</math> can be factored as <math>(x+1)(x-1)</math>. The <math>x-1</math> terms in the numerator and denominator cancel, so the expression is equal to <math>x+1</math> so long as <math>x</math> does not equal <math>1</math>. Looking at the function's behavior near 1, we see that as <math>x</math> approaches one, the expression approaches <math>\boxed{\textbf{(D)}\ 2}</math>. | ||
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| + | Note: Alternatively, one can ignore the domain restriction and just plug in <math>x = 1</math> into the reduced expression. | ||
| + | ~cxsmi | ||
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{{AHSME 50p box|year=1950|num-b=39|num-a=41}} | {{AHSME 50p box|year=1950|num-b=39|num-a=41}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:10, 19 January 2024
Problem
The limit of 
as 
approaches 
as a limit is:
Solution
Both 
and 
approach 0 as approaches , using the L'Hôpital's rule, we have 
.
Thus, the answer is 
.
~ MATH__is__FUN
Solution 2
The numerator of can be factored as 
. The terms in the numerator and denominator cancel, so the expression is equal to 
so long as does not equal . Looking at the function's behavior near 1, we see that as approaches one, the expression approaches .
Note: Alternatively, one can ignore the domain restriction and just plug in 
into the reduced expression.
~cxsmi
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 39 |
Followed by Problem 41 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
