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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.
A function is a relation for which each of the value from the set the first components of the ordered pairs is related with exactly one value from the set of second components of t
[3+tan20+tan80]/tan20+tan80
If e were rational, then e = n/m for some positive integers m, n. So then 1/e = m/n. But the series expansion for 1/e is 1/e = 1 - 1/1! + 1/2! - 1/3! + ... Call the first n v
The last topic that we want to discuss in this section is that of intercepts. Notice that the graph in the above instance crosses the x-axis in two places & the y-axis in one plac
Divergence Test Once again, do NOT misuse this test. This test only says that a series is definite to diverge if the series terms do not go to zero in the limit. If the
A. Design an investigation that details the following six components:
2/4t=1/2
find a common tangent to two circles
I need help solving principal equations where interest,rate,and time are given.
The subsequent topic that we require to take a look at is the determinant of a matrix. The determinant is in fact a function that gets a square matrix and converts this in a number
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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