Line, Point, Curve
f ( x ) = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ( x 2 − 4 ) / ( x − 2 ) , 2 , x 3 − 3 x 2 + 2 x + 4 , if x < 2 if x = 2 if x > 2
Compute x → 2 lim f ( x ) .
The answer is 4.
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2 solutions
At x = 2 there is a removable discontinuity, so the limit is well defined. That is, the function is continuous at all points near x = 2 except at that precise point. Because of this case the limit exists.
The solution isn't 4. Even the limit of f(x), when x tends to (2^-) seems to be equal to the limit of f(x) when x tends to (2^+) that is 4. The rule to make sure that a function has a limit when x tends to y, is that:f(y) =( limit/x -->(y^- or y^+)) (f(y)). So, as f(2) =2 ≠of ( limit/x -->(2^- or 2^+)) (f(x))=4. The conclusion we came to is that does not exist the limit of f(x) when the x tends to 2.
If I am mistaken, please correct me and tell your point of view.
In order for a limit to exist, the limit from the left (negative) and right (positive) must be equal. From the first domain, when x approaches 2 from the left, x → 2 − lim f ( x ) = x → 2 − lim ( x − 2 x 2 − 4 ) = x → 2 − lim x + 2 = 4 .
In other words, even though the rational function can't really be defined at x = 2 , we know that the limit is at the point (2 , 4) .
Similarly, when x approaches 2 from the right, x → 2 + lim f ( x ) = x → 2 + lim x 3 − 3 x 2 + 2 x + 4 = 4 .
Again, even though f ( 2 ) equals to 2 in this case, but the limit from the right does not. Instead, the curvy graph will approach the point (2 , 4) from the right side.
Hence, since the limits from both sides are equal, the limit at x = 2 is defined, and x → 2 lim f ( x ) = 4 .
If I am right, limit at a point exists iff: x → 0 + lim = x → 0 − lim = f ( 0 ) Shouldn't then the unswer be undefined... Any ways please explain..
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The value of f(x) itself isn't necessarily included in the limit existence though if such limit equals to f(x), we can say it's a continuous function.
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Right.. Thanks.... I confused it with continuity...
how did u get to lim x-2=4 from the one before it?
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Simple factor division. x 2 − 4 = ( x − 2 ) ( x + 2 ) .
If you get the limit to be something in an indeterminate form like 0/0 you can use L'Hopital's rule. You can differentiate the numerator and denominator and try subbing in again. https://en.m.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
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But wouldn't i get 2x/x by that wich would be equal to 2 Independent of x?
what if both side are not equal?
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Should the limits from both sides are not equal at a specific value of x, the limit at that value will not exist.
good solution. There is a small error, in the fifth line should be 'right' instead of left. Can we conclude that our function is discontinuous in f(2), since lim x-->2 f(x) is different from f(2)?
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Thank you for your correction, and yes, we can say that it's discontinuous at x=2 despite its defined value.
This is a discontinuous function. So it seems to me the idea of a having a limit at that point is a bit of a red herring.