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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.
An $80.00 coat is marked down 20%. It does not sell, so the shop owner marks it down an additional 15%. What is the new price of the coat? Find out 20 percent of the original p
If the population standard deviation is o=8, how large a sample is necessary to have a standard error that is: a. less than 4 points? b. less than 2 points? c. less than 1 poin
Explain the Dependent Events? Events are called dependent events when the outcome of one event influences the outcome of the second event. P(A and B) = P(A) P(B following A
Let Consider R A Χ B, S B Χ C be two relations. Then compositions of the relations S and R given by SoR A Χ C and is explained by (a, c) €(S o R) iff € b € B like (a, b) € R,
solution for this project
if you have 1/5 of a candy bar and 4 friends how much will they get
what is dot
2(sin 6 ?+cos 6 ?) - 3(sin 4 ?+cos 4 ?)+1 = 0 Ans: (Sin 2 ?)3 + (Cos 2 ?)3-3 (Sin 4 ?+(Cos 4 ?)+1=0 Consider (Sin 2 ?)3 +(Cos 2 ?)3 ⇒(Sin 2 ?+Cos 2 ?)3-3 Sin 2 ?Co
Geometric Applications to the Cross Product There are a so many geometric applications to the cross product also. Assume we have three vectors a → , b → and c → and we make
Use the graph of y = x2 - 6x to answer the following: a) Without solving the equation (or factoring), determine the solutions to the equation x 2 - 6x = 0 usi
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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