Difference between revisions of "1950 AHSME Problems/Problem 40"

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==Solution==
 
==Solution==
Limits do not take the value of the limiting function at the specified value into account, so we are essentially being asked to find the limit of <math>x+1</math> as <math>x</math> approaches <math>1</math>. This is simply <math>\boxed{\textbf{(D)}\ 2}</math>.
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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>.
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Thus, the answer is <math>\boxed{\textbf{(D)}\ 2}</math>.
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~ MATH__is__FUN
  
 
==See Also==
 
==See Also==

Revision as of 06:53, 3 December 2023

Problem

The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$

Solution

Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$, using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$. Thus, the answer is $\boxed{\textbf{(D)}\ 2}$.

~ MATH__is__FUN

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 41
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All AHSME Problems and Solutions

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