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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.
Ask Suppose I offer you a loan to start a safety matchstick production unit on the following terms: I shall first advance you Rs.50,000/- to set up your unit, and wait for 3 month
theory about solving sequencing problem using graphical method
#question.what is the meaning of this
How do I increase and decrease tax and sales
What is Pythagorean Triples? A set of three numbers a, b, and c that can satisfy the equation A 2 +b 2 = c 2 , is called a Pythagorean triple. The following is a list of
solution for this project
Calculate the area of RECTANGLE ? The area of a rectangle is the amount of space taken up by a rectangle, which is a two-dimensional shape. You find the area (A) of a recta
At time t an investor shorts a $1 face value zero coupon bond that matures at time T = t and uses the entire proceeds to purchase a zero coupon bond that matures at time
(1 0 3 21 -1 1 -1 1) find A-1
1. Calculate the annual interest that you will receive on the described bond-A $500 Treasury bond with a current yield of 4 .2% that is quoted at 106 points? 2. Compute the tota
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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