Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

For the function f(x)f(x) shown below, determine \displaystyle \lim_{x\rightarrow 2}f(x).x→2lim​f(x).f(x)=\begin{cases}-1+x^2 & \text{ for }\hspace{10px} x<2\phantom{\frac{1}{1}} \\ -1+x & \text{ for }\hspace{10px} x\ge2\phantom{\frac{1}{1}}\end{cases}f(x)={−1+x2−1+x​for x<211​for x≥211​​

Sagot :

Using lateral limits, it is found that the limit of the function does not exist.

The function is defined by parts, that is, it's rule depends on the input. The function is given by:

[tex]f(x) = -1 + x^2, x < 2[/tex]

[tex]f(x) = -1 + x, x \geq 2[/tex]

What is a limit?

A limit is given by the value of function f(x) as x tends to a value. If the function is piece-wise, that is, it has multiple definitions, at the point of which the values of x changes, lateral limits have to be calculated.

The lateral limits are given by:

[tex]\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2} -1 + x^2 = -1 + 2^2 = 3[/tex]

[tex]\lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2} -1 + x = -1 + 2 = 1[/tex]

Since the lateral limits at x = 2 are different, the limit of f(x) as x goes to 2 does not exist.

You can learn more about lateral limits at https://brainly.com/question/23343679