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lim x->3 1/(x-3)
First substitute 3 in for x, this gives 1/0 which is undefined and not allowed
So, next set up limits from the left and right sides
From the left
lim x->3- 1/(x-3) (the - is the symbol for from the left), this gives 1/a very small neg
1/a very small neg= -infinity
From the right
lim x->3+ 1/(x-3), this gives 1/a very small positive number
1/small pos=infinity
The left should equal the right; since it does not, there is an asymptote at x=3
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What is the limit as x approaches 1 when 1 minus square root of 2x squared minus 1 divided by x-1?
The answer will depend on any parentheses present in the expression. Until these are given explicitly, it is not possible to answer the question.
Is zero divided by zero equal to zero?
Actually 0/0 is undefined because there is no logical way to define it. In ordinary mathematics, you cannot divide by zero.The limit of x/x as x approaches 0 exists and equals 1 so you might be tempted to define 0/0 to be 1.However, the limit of x2/x as x approaches 0 is 0, and the limit of x/x2 as x approaches 0 does not exist .r/0 where r is not 0 is also undefined. It is certainly misleading, if not incorrect to say that r/0 = infinity.If r > 0 then the limit of r/x as x approaches 0 from the right is plus infinity which means the expression increases without bounds. However, the limit as x approaches 0 from the left is minus infinity.
What is the limit of sin multiplied by x minus 1 over x squared plus 2 as x approaches infinity?
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
How do you solve the limit 1 minus cos3x divided by sin3x as approaches 0?
Use l'Hospital's rule: If a fraction becomes 0/0 at the limit (which this one does), then the limit of the fraction is equal to the limit of (derivative of the numerator) / (derivative of the denominator) . In this case, that new fraction is sin(3x)/cos(3x) . That's just tan(3x), which goes quietly and nicely to zero as x ---> 0 . Can't say why l'Hospital's rule stuck with me all these years. But when it works, like on this one, you can't beat it.
What is E raised to the power minus infinite?
You can't really take that power, but you can take the limit - meaning you can see what happens when the exponent gets closer and closer to "minus infinite". That limit is zero.
What is the limit as x approaches 1 when 1 minus square root of 2x squared minus 1 divided by x-1?
The answer will depend on any parentheses present in the expression. Until these are given explicitly, it is not possible to answer the question.
Is zero divided by zero equal to zero?
Actually 0/0 is undefined because there is no logical way to define it. In ordinary mathematics, you cannot divide by zero.The limit of x/x as x approaches 0 exists and equals 1 so you might be tempted to define 0/0 to be 1.However, the limit of x2/x as x approaches 0 is 0, and the limit of x/x2 as x approaches 0 does not exist .r/0 where r is not 0 is also undefined. It is certainly misleading, if not incorrect to say that r/0 = infinity.If r > 0 then the limit of r/x as x approaches 0 from the right is plus infinity which means the expression increases without bounds. However, the limit as x approaches 0 from the left is minus infinity.
What is the limit of sin multiplied by x minus 1 over x squared plus 2 as x approaches infinity?
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
How do you solve the limit 1 minus cos3x divided by sin3x as approaches 0?
Use l'Hospital's rule: If a fraction becomes 0/0 at the limit (which this one does), then the limit of the fraction is equal to the limit of (derivative of the numerator) / (derivative of the denominator) . In this case, that new fraction is sin(3x)/cos(3x) . That's just tan(3x), which goes quietly and nicely to zero as x ---> 0 . Can't say why l'Hospital's rule stuck with me all these years. But when it works, like on this one, you can't beat it.
What is E raised to the power minus infinite?
You can't really take that power, but you can take the limit - meaning you can see what happens when the exponent gets closer and closer to "minus infinite". That limit is zero.
What is the limit of x squared plus four x plus four all over x squared minus 4 as x approaches 2?
The "value" of the function at x = 2 is (x+2)/(x-2) so the answer is plus or minus infinity depending on whether x approaches 2 from >2 or <2, respectively.
What is the limit as x approaches 2 of square root of x plus 1 minus 3 divided by x minus 2?
It is not possible to answer the question because it is not clear whether the expression you have in mind is:sqrt(x + 1) - 3/x - 2or sqrt[x + 1 - 3/x - 2]or sqrt(x + 1) - 3/(x - 2)or sqrt[x + 1 - 3/(x - 2)]or some other variant.
Limit x approaches 1 x minus 1 divided by x cubed minus 1?
limx->1 (x-1)/(x^3-1). You can factor and solve it that way, but I'm going to show a cooler way to do it: If you've heard of L'Hopital's Rule, then you know that limx->1 (x-1)/(x^3-1)=limx->1 (1/3x^2)=1/3(1)=1/3
Limit as t approches 4 square t minus 2 divided by t-4?
Since both the numerator and the denominator approach zero, the conditions for de l'Hospital's rule are fulfilled. Take the derivate of both the numerator and the denominator, and take the limit of the new fraction: lim (t2-2)/(t-4) = lim 2t / 1 = 8 / 1 = 8.
Patricia a standard card holder uses her card for the first time and makes a 500 cash withdrawal How much of her credit limit remains?
This question does not supply enough information to be fully answered. It would depend on her contract and card limit. If her card limit was $1,000, her remaining limit would be $500. If her card limit was $700, her remaining limit would be $200. You would take whatever her total limit is, and subtract the amount she has withdrawn or used. If this were a debit card instead of a credit card, it would be however much she had deposited, minus the amount she has used or withdrawn. Some debit accounts require a minimum balance be kept in the account at all times.
How pH can be negative?
When the concentration exceed 10mol per cubic decimeter. pH equals minus log[H+]
What is the solution for limit x tends to zero tank minus sink whole divided by k?
Since x is not a part of the expression, x can approach zero without any effect. So, the answer would be (tank-sink whole)/k, k<>0.
