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As x tends to zero the value of 1/x tends to either ∞ or -∞. In this situation we will not be sure about the exact value of 1/x. As a result we will not be sure about the exact/approaching value of sin(1/x). We cant say anything about the value of sine function unless we know the angle and in this question we are not sure about the angle as at infinity it can take any value. We will be sure that the value of sin(1/x) will lie in [-1, 1] but not sure about a unique value. As in limits, it exists only when we get a unique value. Therefore we will say that the limit does not exist.
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Kwai made 5 pints of iced tea. How many cups of tea did he make?
143
Let f : R 3 → R be de?ned by: f(x, y, z) = xy 2 + x 3 z 4 + y 5 z 6 a) Compute ~ ∇f(x, y, z) , and evaluate ~ ∇f(2, 1, 1) . b) Brie?y
Evaluate the area and perimeter of a square: Example: Calculate the area and perimeter of a square with a = 5´. Be sure to include units in your answer. Solution:
We know that 2 4 = 16 and also that 2 is referred to as the base, 4 as the index or power or the exponent. The same if expressed in terms of logarithms would be log 2
20+20
3 3/4+(1 1/49*7/10)
examples of least cost method
It is the catch all force. If there are some other forces which we decide we need to act on our object we lump them in now and call this good. We classically call F(t) the forcing
Limit sin(1/x) when x tends to 0 is not definedCan be proved simply by multiplying and dividing by x then xsin(1/x)/x becomes 1/x as xsin(1/x)or for that matter sin(1/x)/1/x = 1 and limit reduces to 1/x which doesnt exist Also the proof can be that when x approcashes 0 from positive side 1/x tends to positive infinty and limit (right0 becomes sin(infinity) but when from left side 1/x tends to negative infinty so limit becomes -sin(infinit) which both can never b equal. so limit doesnt exist
Limit sin(1/x) when x tends to 0 is not definedCan be proved simply by multiplying and dividing by x then xsin(1/x)/x becomes 1/x as xsin(1/x)or for that matter sin(1/x)/1/x = 1 and limit reduces to 1/x which doesnt exist Also the proof can be that when x approcashes 0 from positive side 1/x tends to positive infinty and limit (right0 becomes sin(infinity) but when from left side 1/x tends to negative infinty so limit becomes -sin(infinit) which both can never b equal.
so limit doesnt exist
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