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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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: Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both 3 inches.
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Ashley's car insurance costs her $115 per month. How much does it cost her per year? Multiply $115 by 12 because there are 12 months in a year; $115 × $12 = $1,380 per year.
Need assignment help, Explain Multiplication of two Matrices.
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Mathematical Problem Solving In 1945, mathematician George Polya (1887-1985) published a book titled How To Solve It in which he demonstrated his approach to solving problems.
Array - when items are arranged in a regular rectangular pattern of rows and columns, counting how many there are. (e.g., if there are 3 rows of 5 girls each, how many girls are t
limit 0 to 2(3x^2+2) Solution) integrate 3x^2 to x^3 and 2 to 2x and apply the limit from 0 to 2 answer is 12.
Explain Equivalent Fractions ? Two fractions can look different and still be equal. Different fractions that represent the same amount are called equivalent fractions. Ar
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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