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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.
In pharmaceutical product research doctors visit the place to learn what
a group of 3o students is planning a thanksgiving party items needed hats @ $2.50 each.noise makers@$4.00 per pack of 5.Ballons @$5.00 per pack of 10.how many packs of noisemakers
A Class 4 teacher was going to teach her class fractions. At the beginning of the term she asked the children, "If you had three chocolates, and wanted to divide them equally among
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In class 1, the teacher had written down the digits 0,1, ...., 9 on the board. Then she made all the children recite the corresponding number names. Finally, she made them write th
Position Vector There is one presentation of a vector that is unique in some way. The presentation of the ¯v = (a 1 ,a 2 ,a 3 ) that begins at the point A = (0,0,0) and ends
log2(x^2)=(log2(x))2
What are the characteristics of a queuing system? (i) The input pattern (ii) The queue discipline (iii) The service mechanism
My nephew had been introduced to division by his teacher Ms. Santosh, in Class 3. He, and several of his friends who had been taught by her, appeared to be quite comfortable with t
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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