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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The answer will depend on any parentheses present in the expression. Until these are given explicitly, it is not possible to answer the question.
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.
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
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.
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.