1
The limit is 0.
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
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.
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.
The limit is 1.
The limit is 0.
2
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
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.
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.
The limit is 1.
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.
One over A squared or A to the negative 2.
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.
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.
2
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