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Find the indicated limit, if it exists.

limit of f of x as x approaches 2 where f of x equals x plus 3 when x is less than 2 and f of x equals 3 minus x when x is greater than or equal to 2

The limit does not exist.
1
2
5

Find the indicated limit, if it exists. limit of f of x as x approaches 2 where f of x equals x plus 3 when x is less than 2 and f of x equals 3 minus x when x class=

Answer :

LammettHash

[tex]\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}(x+3)=5[/tex]

[tex]\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}(3-x)=1[/tex]

The limits from either side don't match, so the two-sided limit does not exist.

The limit does not exist for the function [tex]f(x) = \left \{ {x+3, {x < 2} \atop 3-x,{x\geq 2}} \right.[/tex]  at x=2. This can be obtained by calculating the left hand limit (LHL) and right hand limit (RHL) and checking whether they are equal.

What is the rule to compute LHL and RHL?

  • LHL, put x=a-h, h>0 and then take limit as h→0⁻

LHL = [tex]\lim_{h \to \ 0 } f(a-h)[/tex]

  • RHL, put x=a+h, h>0 and then take limit as h→0⁺

RHL = [tex]\lim_{h \to \ 0 } f(a+h)[/tex]

Calculate the LHL and RHL:

  • LHL of f(x) at x = 2 is

[tex]\lim_{x \to \ 2^{-} } f(x)[/tex] = [tex]\lim_{h \to \ 0} f(2-h)[/tex]

                     = [tex]\lim_{h \to \ 0} (2-h+3)[/tex]

                     = [tex]\lim_{h \to \ 0} (5-h)[/tex]

                     = 5

  • RHL of f(x) at x = 2 is

[tex]\lim_{x \to \ 2^{+} } f(x)[/tex] = [tex]\lim_{h \to \ 0} f(2+h)[/tex]

                     = [tex]\lim_{h \to \ 0 } (3-2+h)[/tex]

                     = [tex]\lim_{h \to \ 0 } (1+h)[/tex]

                     = 1

Clearly LHL≠RHL

Therefore the limit does not exist.

Hence the limit does not exist for the function [tex]f(x) = \left \{ {x+3, {x < 2} \atop 3-x,{x\geq 2}} \right.[/tex]  at x=2.

Learn more about limits here:

brainly.com/question/14344608

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