Use the given graph to determine the limit, if it exists. A coordinate graph is shown with a horizontal line crossing the y axis at six that ends at the open point 2, 6, a closed point at 2, 1, and another horizontal line starting at the open point 2, negative 3. Find limit as x approaches two from the left of f of x. and limit as x approaches two from the right of f of x..

as the graph moves from the left, over and over and over is consistently 6
when x = 1, limit is 6, 1.5, 6, 1.9, still 6, 1.999999999 still 6, 1.9999999999999999999999999 the limit is 6 still, now when x = 2, y = 1

but for the sake of a limit, it doesn't really matter if y = 6 or not at x = 2

as "x" is moving over and over and over from the left, the limit is consistenly, 6
now, it will not be 6 at x = 2, but that doesn't matter
the limit is "what is 'y' approaching as 'x' moves along", and is 6 from the left

now, from the right, is consistenly -3
same, as x = 2, y = 1 but
at x = 3, the limit is -3, x = 2.9, -3 still, 2.1, -3 still, 2.00000000000000000000000000000000001, the limit is still -3
thus, from the right, is -3

DelcieRiveria DelcieRiveria · Answer 2 · 2018-12-06T06:28:04-05:00

Answer:

The limit of the function does not exists.

Step-by-step explanation:

From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.

For x<2

\(f(x)=6\)

The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.

For x>2

\(f(x)=-3\)

The value of y is 1 at x=2, because of he close circles on (2,1).

For x=2

\(f(x)=1\)

Therefore the graph represents a piecewise function, which is defined as

\(f(x)=\begin{cases}6& \text{ if } x<2\\ 1& \text{ if } x=2 \\ -3& \text{ if } x>2 \end{cases}\)

The limit of a function exist at a point a if the left hand limit and right hand limit are equal.

\(lim_{x\rightarrow a^-}f(x)=lim_{x\rightarrow a^+}f(x)\)

The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.

\(lim_{x\rightarrow 2^-}f(x)=6\)

\(lim_{x\rightarrow 2^+}f(x)=-3\)

\(6 eq-3\)

Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.