Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.