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The answer is no.
(1)Consider the function given by `f(x)=1/(x-3)` for `x!=3` , and 3 if x=3. The limit as `x->3` does not exist, but the function is defined at 3.
(2) Consider the function `f(x)=x^2` if x>3, `f(x)=x` for `x<=3` . Then the limit as `x->3` does not exist, but `f(3)=3` .
The limit can fail to exist because the function grows without bound or goes to negative infinity. The limit could also fail if there is a jump discontinuity.
The limit of a function fails to exist if the value of the limit of function, as x -> 3 from the left side is not equal to the value of the limit of the function as x-> 3 from the right side.
I suggest to select the function `f(x)=(x-2)/(x-3)(x+1)` and to evaluate its limit as x->3 from the left side such that:
`lim_(x-gt3,xlt3) (x-2)/(x-3)(x+1) = (2.9999...9 - 3)/0 = -oo`
valuating the limit of function as x->3 from the right side yields:
`lim_(x-gt3,xgt3) (x-2)/(x-3)(x+1) = (3.0000000...01 - 3)/0 = +oo`
The limits fails to exist if x->3 since the left side and right side values of limits are different.
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