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What is the limit as x approaches 2 of x² - 2x over x² - x - 2?
Wiki User
∙ 12y ago
Lim(x→2) (x2 - 2x) / (x2 - x - 2)
= Lim(x→2) x(x - 2) / (x - 2)(x + 1)
= Lim(x→2) x / (x + 1)
= 2/3
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∙ 12y ago
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What is the limit as x approaches 2 of 1x-22?
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How do you integrate 2x over 2x-2?
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2x * 2x (2*2=4) therefore, 2x*2x equals 4x2.
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2x + 2 is an algebraic expression. It is also a polynomial and it has two terms: 2x and 2.
What is negative 2 to the fifth power?
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What is the limit of 1-cos x over x squared when x approaches 0?
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How do you solve the limit as x approaches 90 degrees of cos 2x divided by tan 3x?
Take the limit of the top and the limit of the bottom. The limit as x approaches cos(2*90°) is cos(180°), which is -1. Now, take the limit as x approaches 90° of tan(3x). You might need a graph of tan(x) to see the limit. The limit as x approaches tan(3*90°) = the limit as x approaches tan(270°). This limit does not exist, so we'll need to take the limit from each side. The limit from the left is ∞, and the limit from the right is -∞. Putting the top and bottom limits back together results in the limit from the left as x approaches 90° of cos(2x)/tan(3x) being -1/∞, and the limit from the right being -1/-∞. -1 divided by a infinitely large number is 0, so the limit from the left is 0. -1 divided by an infinitely large negative number is also zero, so the limit from the right is also 0. Since the limits from the left and right match and are both 0, the limit as x approaches 90° of cos(2x)/tan(3x) is 0.
What is the limit of sin multiplied by x minus 1 over x squared plus 2 as x approaches infinity?
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Why does 7x over 2 equal 2x plus x plus x over 2?
7x = 4x+2x+1xSo 7x/2 = 4x/2 + 2x/2 + x/2 (by the distributive property of division over addition) = 2x + x + x/2
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What is the limit of sine squared x over x as x approaches zero?
So, we want the limit of (sin2(x))/x as x approaches 0. We can use L'Hopital's Rule: If you haven't learned derivatives yet, please send me a message and I will both provide you with a different way to solve this problem and teach you derivatives! Using L'Hopital's Rule yields: the limit of (sin2(x))/x as x approaches 0=the limit of (2sinxcosx)/1 as x approaches zero. Plugging in, we, get that the limit is 2sin(0)cos(0)/1=2(0)(1)=0. So the original limit in question is zero.
Does limit always tend to zero only?
No, limit can tend to any finite number including 0. It is also possible that the limit does not tend to any finite value or approaches infinity. Example: The limit of x^2+5 tend to 6 as x approaches -1.
What is the limit of x squared plus four x plus four all over x squared minus 4 as x approaches 2?
The "value" of the function at x = 2 is (x+2)/(x-2) so the answer is plus or minus infinity depending on whether x approaches 2 from >2 or <2, respectively.
