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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.
Consider the following linear equations. x1-3x2+x3+x4-x5=8 -2x1+6x2+x3-2x4-4x5=-1 3x1-9x2+8x3+4x4-13x5=49
Example 1 Add 4x 4 + 3x 3 - x 2 + x + 6 and -7x 4 - 3x 3 + 8x 2 + 8x - 4 We write them one below the other as shown below.
Question: Classify the following differential equations as linear/nonlinear. Also, what is the order of the following differential equations? Xy'-2y =x Xy'' -2y' =xsin(y)
Solve 2 ln (√x) - ln (1 - x ) = 2 . Solution: Firstly get the two logarithms combined in a single logarithm. 2 ln (√x) - ln (x - l) = 2 ln ((√x) 2 ) ln (1 - x ) = 2
The Mean Value Theorem Assume f(x) is a function that satisfies both of the subsequent. 1. f(x) is continuous on the closed interval [a,b]. 2. f(x) is differentiabl
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how many words can be formed from letters of word daughter such that each word contain 2vowles and 3consonant
Consider the trigonometric function f(t) = -3 + 4 cos(Π/ 3 (t - 3/2 )). (a) What is the amplitude of f (t)? (b) What is the period of f(t)? (c) What are the maximum and mi
GIVE EXAMPLE OF ROW EQUIVALENT
solutions
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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