Difference between revisions of "1950 AHSME Problems/Problem 40"
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==Solution== | ==Solution== | ||
| − | + | Both <math>x^2-1</math> and <math>x-1</math> approach 0 as <math>x</math> approaches <math>1</math>, using the L'Hôpital's rule, we have <math>\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2</math>. | |
| + | Thus, the answer is <math>\boxed{\textbf{(D)}\ 2}</math>. | ||
| + | |||
| + | ~ MATH__is__FUN | ||
==See Also== | ==See Also== | ||
Revision as of 07:53, 3 December 2023
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
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 39 |
Followed by Problem 41 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
