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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.
Which of the subsequent numbers will yield a number larger than 23.4 while it is multiplied by 23.4? When multiplying through a number less than 1, you get a product in which i
Inverse Functions : In the last instance from the previous section we looked at the two functions f ( x ) = 3x - 2 and g ( x ) = x /3+ 2/3 and saw that ( f o g ) ( x )
explane
how do you graph y+3=-x+3x on a TI-83 graphing calculator?
Two people are 50 feet separately. One of them begin walking north at rate so that the angle illustrated in the diagram below is changing at constant rate of 0.01 rad/min. At what
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Shane rolls a die numbered 1 by 6. What is the probability Shane rolls a 5? From 2:15 P.M. to 4:15 P.M. is 2 hours. After that, from 4:15 P.M. to 4:45 P.M. is another half hour
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What is the exact vale of sin(theta/2) when sintheta=3/5, pi/2
Two circles touch externally. The sum of their areas is 58 π cm 2 and the distance between their centres is 10 cm. Find the radii of the two circles. (Ans:7cm, 3cm) Ans:
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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