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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.
Write Prim's Algorithm. Ans: Prim's algorithm to find out a minimum spanning tree from a weighted graph in step by step form is given below. Let G = (V, E) be graph and S
What are the factors of 956
Evaluate the given definite integral. Solution Let's begin looking at the first way of dealing along with the evaluation step. We'll have to be c
The alternative hypothesis When formulating a null hypothesis we also consider the fact that the belief may be found to be untrue thus we will refuse it. Therefore we formula
The perimeter of Andrew''s rectangular room is 44 feet. What equation was used to find the perimeter?
compare 643,251;633,512; and 633,893. The answer is 633,512
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Here, we have tried to present some of the different thinking and learning processes of preschool and primary school children, in the context of mathematics learning. We have speci
112 in 8
From top of a tower a stone is thrown up and it reaches the ground in time t1. A second stone is thrown down with the same speed and it reaches the ground in t2. A third stone is r
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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