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What is the limit of x cosine 1 over x squared as x approaches 0?
Wiki User
∙ 9y ago
The limit is 0.
Wiki User
∙ 9y ago
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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 1-cos x over x squared when x approaches 0?
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.
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 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 1-cos x over x squared when x approaches 0?
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.
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.
How do you prove that the sin over one minus the cosine minus one plus the cosine over the sine equals zero?
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.
Why do cosine to the negative 1th power and cosine cancel out each other?
Cosine to the negative first power and cosine cancel each other out because cosine to the negative first power is one over cosine, and one over anything times anything is just one.
What is cos squared of pi over 4?
If working in radians, the cosine of one-quarter of pi is equal to the square root of 0.5. Squaring this gives 0.5, or exactly one-half.
What number has a cosine of 15 over 17?
The cosine of 0.489957 radians is 15/17
What the cosine of 45 degrees?
cosine 45° = √2/2 (Square root of 2 over 2)
Why does sec 0 equals 1?
Secant is 1 over cosine and cosine 0 equals 1.
