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

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==Problem==
 
==Problem==
  
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The limit of <math> \frac {x^2\minus{}1}{x\minus{}1}</math> as <math>x</math> approaches <math>1</math> as a limit is:
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The limit of <math> \frac {x^2-1}{x-1}</math> as <math>x</math> approaches <math>1</math> as a limit is:
  
 
<math>\textbf{(A)}\ 0 \qquad
 
<math>\textbf{(A)}\ 0 \qquad
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==Solution==
 
==Solution==
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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
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==Solution 2==
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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, we can ignore the domain restriction and just plug in <math>x = 1</math> into the reduced expression.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
  
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==See Also==
 
 
{{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]]
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{{MAA Notice}}

Latest revision as of 12:43, 4 April 2024

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

Solution 2

The numerator of $\frac {x^2-1}{x-1}$ can be factored as $(x+1)(x-1)$. The $x-1$ terms in the numerator and denominator cancel, so the expression is equal to $x+1$ so long as $x$ does not equal $1$. Looking at the function's behavior near 1, we see that as $x$ approaches one, the expression approaches $\boxed{\textbf{(D)}\ 2}$.


Note: Alternatively, we can ignore the domain restriction and just plug in $x = 1$ into the reduced expression.

~ cxsmi

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Problem 41
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