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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.
Imagine a time in history when the number system had not yet evolved a farmer needed to keep track of his cattle. What would he do to figure out whether his entire rattle returned
If 4x^4+9x^4=64 then the maximum value of x^2+y^2 is solution) From the eq. finding the value of x^2 and putting it in x^2 + y^2.we get 2nd eq. differentiating that and putting
Sketch the phase portrait for the given system. Solution : From the last illustration we know that the eigenvectors and eigenvalues for this system are, This tu
2/4t=1/2
What is Inductive Reasoning ? Sometimes we draw conclusions based on our observations. If we observe the same results again and again, we conclude that the event always has the
my math skills are keeping me from getting my ged need help in all areas
What is Slope of a Line ? A line can have a "steep" slope or a "gradual" slope. slope = rise/run The "rise" is the distance going up or down. The "run" is the distance goin
It is the last case that we require to take a look at. During this section we are going to look at solutions to the system, x?' = A x? Here the eigenvalues are repeated eigen
Go back to the complex numbers code in Figures 50 and 51 of your notes. Add code fragments to handle the following: 1. A function for adding two complex numbers given in algeb
#question.Mai is 3 years ypunger than twice the age of her brother .If b represents .
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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