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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.
#question.onstruct/draw geometric shapes with specific condition.
How do you find the second minimum spanning tree of a graph? Find the second minimum spanning tree of the following graph. Ans: The second minimum spanning tree is acq
a man can row a bangka at a rate of 5 km/h in still water. It takes 10 minutes longer to row upstream a distance of 2km than he takes to row downstream. What is the rate of the cur
what is 5x - 14x + 7x =
can you please help me with this topic that im on in classand I just don''t get it and can u help me with dividing fractions adding mutply subtract add
a car comes to a stop from a speed of 30m/s in a distance of 804m. The driver brakes so as to produce a decelration of 1/2m per sec sqaured to begin withand then brakes harder to p
This time we are going to take a look at an application of second order differential equations. It's now time take a look at mechanical vibrations. In exactly we are going to look
the function g is defined as g:x 7-4x find the number k such that kf(-8)=f- 3/2
how to round off 42,999,523
x ?8x ?16x ? 0, x?0? ? 0, x?0? ? 2
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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