document.write( "Question 1000949: 1. let f(x) = ((2x-1)^2)/(2x^2)
\n" ); document.write( "find:
\n" ); document.write( "a. lim x->∞
\n" ); document.write( "b. lim x->-∞
\n" ); document.write( "c. lim x->0 \r
\n" ); document.write( "\n" ); document.write( "work:
\n" ); document.write( "a. lim->∞
\n" ); document.write( "((2(∞)-1)^2)/(2(∞)^2) = (4(∞)^2)/(2(∞)^2) = 2. Because we dismiss the -1 and the (∞)^2 cancel eachother out. \r
\n" ); document.write( "\n" ); document.write( "b. lim x->-∞
\n" ); document.write( "(2(-∞)-1^2)/(2(-∞)^2) = 2. For same reasons. \r
\n" ); document.write( "\n" ); document.write( "c. lim x->0
\n" ); document.write( "((2(0)-1)^2)/(2(0)) = ((-1)^2)/(0) = DNE because you cannot divide by zero therefore a two-sided limit diverges at zero. \r
\n" ); document.write( "\n" ); document.write( "============================ \r
\n" ); document.write( "\n" ); document.write( "Similar problem:
\n" ); document.write( "Let f(x) = ((2x-2)^2)/((x-3)^2)
\n" ); document.write( "find:
\n" ); document.write( "a. lim x->∞
\n" ); document.write( "b. lim x->-∞
\n" ); document.write( "c. lim x->3^(+)
\n" ); document.write( "d. lim x->3^(-) \r
\n" ); document.write( "\n" ); document.write( "work:
\n" ); document.write( "a. lim x->∞
\n" ); document.write( "((2(∞)-2)^2)/((∞-3)^2)
\n" ); document.write( "((2(∞))^2)/((∞)^2))
\n" ); document.write( "(4(∞)^2)/(∞)^2 = 4 because (∞)^2's cancel \r
\n" ); document.write( "\n" ); document.write( "b. lim x->-∞
\n" ); document.write( "similarly, this becomes 4 \r
\n" ); document.write( "\n" ); document.write( "c. lim ->3^(+)
\n" ); document.write( "d. lim ->3^(-) \r
\n" ); document.write( "\n" ); document.write( "Bit confused on how to solve these last two. The way I learned it was that 3^(+) approaches 3 from the right, therefore its values is something like 3.00001. And 3^(-) approaches 3 from the left so its value is something like 2.99999. So do I plug and chug these values into the function? or is there another process. One-sided limits confuse me. \r
\n" ); document.write( "\n" ); document.write( "Thank you \n" ); document.write( "
Algebra.Com's Answer #618641 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Maybe it's just semantics but I was taught differently.
\n" ); document.write( "You divide numerator and denominator by the highest order of x and then take the limit.
\n" ); document.write( "\"f%28x%29=%282x-1%29%5E2%2F%28x-3%29%5E2\"
\n" ); document.write( "\"f%28x%29=%284x%5E2-4x%2B1%29%2F%28x%5E2-6x%2B9%29\"
\n" ); document.write( "\"f%28x%29=%284-4%2Fx%2B1%2Fx%5E2%29%2F%281-6%2Fx%2B9%2Fx%5E2%29\"
\n" ); document.write( "Now when you take the limits,
\n" ); document.write( "a)\"lim%28x-%3Einfinity%2Cf%28x%29%29=%284-0%2B0%29%2F%281-0%2B0%29=4\"
\n" ); document.write( "b)\"lim%28x-%3E-infinity%2Cf%28x%29%29=%284-0%2B0%29%2F%281-0%2B0%29=4\"
\n" ); document.write( "c) You can choose values and check.
\n" ); document.write( "\"x=5\"
\n" ); document.write( "\"f%28x%29=%282%285%29-1%29%5E2%2F%285-3%29%5E2=81%2F4\"
\n" ); document.write( "\"x=4\"
\n" ); document.write( "\"f%28x%29=%282%284%29-1%29%5E2%2F%284-3%29%5E2=49\"
\n" ); document.write( "The denominator will get smaller(staying positive), the numerator will get larger(staying positive), so \"f%28x%29-%3Einfinity\"
\n" ); document.write( "Similarly from the left hand side,
\n" ); document.write( "\"x=1\"
\n" ); document.write( "\"f%28x%29=%282%281%29-1%29%5E2%2F%281-3%29%5E2=1%2F4\"
\n" ); document.write( "\"x=2\"
\n" ); document.write( "\"f%28x%29=%282%282%29-1%29%5E2%2F%282-3%29%5E2=9\"
\n" ); document.write( "Similarly, the denominator will get smaller(staying positive), the numerator will get larger(staying positive), so \"f%28x%29-%3Einfinity\"
\n" ); document.write( "So both limits approaching \"x=3\" from the left are positive infinity. \n" ); document.write( "
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