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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
can i get some triangle congruence proofs help?
assuming that the earth''s sphere with a radius of 6400 km.. find the distance along a 3 degree arc at the equator of the earth''s surface?
Determine dy & Δy if y = cos ( x 2 + 1) - x as x changes from x = 2 to x = 2.03 . Solution Firstly let's deetrmine actual the change in y, Δy . Δy = cos (( 2.03) 2
what is 8x6 is
HOW MATHEMATICAL IDEAS GROW : In this section we shall consider three aspects of the nature of mathematical ideas, namely, that they progress from concrete to abstract, from part
44 breaths in 2 hours
Prove: cotA/2.cotB/2.cotC/2 = cotA/2+cotB/2+cotC/2
I need to simple this rational expression, but I can''t figure out how. (x+1)/(x^2-2x-35)+(x^2+x-12)/(x^2-2x-24)(x^2-4x-12)/(x^2+2x-15)
In triangle ABC, cosecA(sinB.sinC+cosB.sinC) is equal to..?
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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