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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.
Determine the second derivative for following functions. Q (t ) = sec (5t ) Solution : Following is the first derivative. Q′ (t
7(y + 3) - 2(x + 2) = 14, 4 (y - 2) + 3(x - 3) = 2 Ans: 7(y + 3) - 2 (x+ 2) = 14 --------- (1) 4(y- 2) + 3(x - 3) = 2 ----------(2) From (1) 7y +21 -
Standard errors of the mean The series of sample means x¯ 1 , x¯ 2 , x¯ 3 ........ is normally distributed or nearly so as according to the central limit theorem. This can be
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Find quadratic equation using the Quadratic Formula: Solve the subsequent quadratic equation using the Quadratic Formula. 4x 2 + 2 = x 2 - 7x: Solution: Step 1.
Intervals which extend indefinitely in both the directions are known as unbounded intervals. These are written with the aid of symbols +∞ and - ∞ . The various types
how to convert double integral into polar coordinates and change the limits of integration
Explain Basic Concepts of Parallel Lines ? Parallel lines are defined in section 1.2 and we use "//" to denote it. From the definition, we can get the following two consequenc
Eileen needs 9 feet of fabric to make a skirt. If Eileen has 18 feet of fabric how many skirts can she make?
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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