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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.
SOLVE THE inequation 0>-5 -X AND X Belongs TO R .Represent THE SOLUTION SET ON THE NUMBER LINE
#question.areas of applications of linear program mes to solution to engineering problems.
performs the mentioned operation and write the answers in standard form. ( -4 + 7 i ) + (5 -10 i ) Solution Actually there isn't much to do here other than add or subt
6x-4+3x+6 -2a-8b-5a+10b
Find all the real solutions to cubic equation x^3 + 4x^2 - 10 =0. Use the cubic equation x^3 + 4x^2 - 10 =0 and perform the following call to the bisection method [0, 1, 30] Use
PLAY AND LEARN : Children can learn many basic mathematical concepts through games. They enjoy Mathematical concepts can be playing within familiar contexts. Their games also gen
what is the differeance in between determinate and matrix .
From past experience a machine is termed to be set up correctly on 90 percent of occasions. If the machine is set up correctly then 95 percent of good parts are expected however i
What is the structure of produt promotion?
E - L - P - S : Has the title of this section stumped you? Children, similarly, don't understand new symbols that are thrust upon them without giving them an adequate grounding. Y
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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