Question

lim x → 3 − f ( x ) = 6 and lim x → 3 + f ( x ) = 5 . As x approaches 3 , f ( x ) approaches 5 , but f ( 3 ) = 6 . As x approaches 3 from the right, f ( x ) approaches 6 . As x approaches 3 from the left, f ( x ) approaches 5 . As x approaches 3 , f ( x ) approaches 6 , but f ( 3 ) = 5 . As x approaches 3 from the left, f ( x ) approaches 6 . As x approaches 3 from the right, f ( x ) approaches 5 . In this situation is it possible that lim x → 3 f ( x ) exists? Explain.

Answer 1

Answer:

Step-by-step explanation:

In limits and continuity of a function, for a function to be continuous, the left hand limit of the function must be equal to the left hand limit and they must be equal to the limit of the function at that point. Mathematically;

$\lim_{n \to a^{+} } f(x) = \lim_{n \to a^{-} } f(x) = \lim_{n \to a^{} } f(x)$

All continuous functions are known to exist but discontinuous functions doesn't exist.

From the question, the point we are considering is at n =3

Given $\lim_{n \to 3^{-} } f(x) =6\ and\ \lim_{n \to 3^{+} } f(x)=5$

but f(3)= 6

As we can see, the left hand limit of the function isn't equal to the left hand limit, this shows that the limit of the function isn't continuous and since all discontinuous function does not exists then  lim x → 3 f ( x ) does not exist.

Answer 2

Limits and continuity of a function:

For only a function to be continuous, its left-hand limitation must be identical to its upper left limit, and both have to be equivalent to the stored procedure limit at a certain point. Simple mathematics;

$\to \lim_{n \to \alpha^+} f(x)= \lim_{n \to \alpha^-} f(x) = \lim_{n \to \alpha} f(x)$

Many continuous functions are known to exist, and none are proven to occur.

According to the query, the point we are examining is at n = 3.

Given:

$\to \lim_{n \to 3^-} f(x)=6 \ \ and \ \ \lim_{n \to 3^+} f(x)=5$

but $f(3)= 6$

As we've seen, the upper left limit of the function is not equivalent to the left hand limit, indicating that the function's restriction is not continuous, and since all discontinuous units need not exist, $\lim_{n \to 3} f(x)$ doesn't exist.

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