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Dereck Kozey

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4y ago

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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 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.


Is infinity over zero an indeterminate form?

Yes, infinity over zero is considered an indeterminate form. This is because while the numerator approaches infinity, the denominator approaches zero, leading to a situation where the expression does not have a well-defined limit. Depending on the context of the limit, the result can vary significantly, making it indeterminate rather than a fixed value.


A squared over A to the fourth?

One over A squared or A to the negative 2.


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.


What is 8x squared over 4x squared?

2


What is 8x squared over 4x squared simplified?

2