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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.
what are multiples?
Leslie ordered a slice of pizza for $1.95, a salad for $2.25, and a soda for $1.05. What was the total cost of her order? The cost of every item must be added together; $1.95 +
2+4
3 divided bye 24
can you explain it to me please
ab=8cm,bc=6cm,ca=5cm draw an incircle.
I need help with my calculus work
Leo works at the Bagel Shop after school and on Saturdays. He is paid $4.00 per hour after school and $5.00 per hour on Saturday. Last week Leo worked a total of 12 hours and made
The geometric mean Merits i. This makes use of all the values described except while x = 0 or negative ii. This is the best measure for industrial increase rates
Basic indefinite integrals The first integral which we'll look at is the integral of a power of x. ∫x n dx = (x n +1 / n + 1)+ c, n
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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