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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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Consider x € R. Then the magnitude of x is known as it's absolute value and in general, shown by |x| and is explained as Since the symbol always shows the nonnegative
If the sides angles of a triangle ABC vary in such a way that it''s circum - radius remain constant. Prove that, da/cos A +db/cos B+dc/cos C=0
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Find the second derivative of the below given equation Y= e x cosx
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if theta is a positive acute angle and 2sin theta +15cos square theta=7 then find the value of cot theta
The Definition of the Limit In this section we will look at the precise, mathematical definition of three types of limits we'll be looking at the precise definition of limits
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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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