0
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.
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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.
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