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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.
How to solve Two-Step Equations? Two-step equations involve two math operations - one operation is addition or subtraction. The second operation is multiplication or division.
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give definition and example
Discontinuous Integrand- Integration Techniques Here now we need to look at the second type of improper integrals that we will be looking at in this section. These are integr
what is meant by "measure of location"
A medical survey was conducted in order to establish the proportion of the population which was infected along with cancer. The results indicated that 40 percent of the population
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(a) An unordered pair fm; ng with 1 ≤ m ≠ n ≤ 6 is called a duad. List the 15 duads. (b) There are 15 ways to partition {1, ......, 6 } into 3 duads, such as { {1; 2}, {3, 4},
Describe Square and Diagonal Matrix.
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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