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What is the limit of 1-cos x over x squared when x approaches 0?
Wiki User
∙ 9y ago
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Dereck Kozey
Wiki User
∙ 9y agoUse L'Hopital's Rule: lim in question =lim of sinx/2x as x approaches 0. Use it again: new limit=cosx/2. So, the limit asked for = cos(0)/2=1/2
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What is the limit of x cosine 1 over x squared as x approaches 0?
The limit is 0.
What is the limit of sin multiplied by x minus 1 over x squared plus 2 as x approaches infinity?
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
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.
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.
What is the Limits of h over tan h as h approaches 0?
The limit is 1.
What is the limit of x cosine 1 over x squared as x approaches 0?
The limit is 0.
What is the limit as x approaches 0 of 1 plus cos squared x over cos x?
2
What is the limit of sin multiplied by x minus 1 over x squared plus 2 as x approaches infinity?
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
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.
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.
What is the Limits of h over tan h as h approaches 0?
The limit is 1.
Why the number 0 has no reciprocal?
Division by zero is not allowed/defined. So you cannot take 'one over zero', or have zero in the denominator.Without going too technical, a person might say that 1/0 is infinity, and it sounds good. But if you have a function [say f(x) = 1/x] and take the limit of f(x) as x approaches zero, then f(x) approaches infinity as x approaches from the right, but it approaches negative infinity as you approach from the left, therefore the limit does not exist.
Evaluate limit x tends to 0 sinx x?
The limit as x approaches zero of sin(x) over x can be determined using the squeeze theorem.For 0 < x < pi/2, sin(x) < x < tan(x)Divide by sin(x), and you get 1 < x/sin(x) < tan(x)/sin(x)That is the same as 1 < x/sin(x) < 1/cos(x)But the limit as x approaches zero of 1/cos(x) is 1,so 1 < x/sin(x) < 1which means that the limit as x approaches zero of x over sin(x) is 1, and that also means the inverse; the limit as x approaches zero of sin(x) over x is 1.You can also solve this using deriviatives...The deriviative d/dxx is 1, at all points. The deriviative d/dxsin(x) at x=0 is also 1.This means you have the division of two functions, sin(x) and x, at a point where their slope is the same, so the limit reduces to 1 over 1, which is 1.
A squared over A to the fourth?
One over A squared or A to the negative 2.
What is 8x squared over 4x squared?
2
What is 8x squared over 4x squared simplified?
2
What is the fraction 25 squared over 36 squared as a decimal?
0.4822530864
