Respuesta :
Answer:
Option 1.
The limit does not exist.
Step-by-step explanation:
The function is
[tex]f(x) = \left \{{{x+3\ \ \ if\ \ \ x<2} \atop{3-x\ \ \ \ if\ \ \ x\geq2}} \right.[/tex]
We seek to find
[tex]\lim_{x \to 2}f(x)[/tex]
Then we must find the limits on the right of 2 ([tex]2 ^+[/tex]) and on the left of 2 ([tex]2 ^ -[/tex])
If both limits exist and are equal then the [tex]\lim_{x \to 2}f(x)[/tex] exists, if both limits are different or do not exist then the [tex]\lim_{x \to 2}f(x)[/tex] does not exist.
Limit on the left of 2
[tex]\lim_{x \to 2^-}x+3 = (2) +3 =5\\\\ \lim_{x \to 2^-}x+3 =5[/tex]
Limit on the rigth of 2
[tex]\lim_{x \to 2^+}3-x = 3-(2) = 1\\\\ \lim_{x \to 2^+}3-x =1[/tex]
Note that both limits give different results. The limit of f(x) when x tends to 2 on the left is equal to 5, and the limit of f(x) when x tends to 2 on the rigth is equal to 1.
Then the [tex]\lim_{x \to 2}f(x)[/tex] does not exist.
Answer:
The correct option is 1.
Step-by-step explanation:
The given function is
[tex]f(x)=\begin{cases}x+3 & \text{ if } x<2 \\ 3-x & \text{ if } x\geq 2 \end{cases}[/tex]
We need to find the [tex]lim_{x\rightarrow 2}f(x)[/tex].
The limit of a function exist if the left hand limit is equal to the right hand limit.
[tex]L=lim_{x\rightarrow a^-}f(x)=lim_{x\rightarrow a^+}f(x)[/tex]
Left hand limit:
[tex]LHL=lim_{x\rightarrow 2^-}f(x)[/tex]
[tex]LHL=lim_{x\rightarrow 2^-}x+3[/tex]
Apply limit.
[tex]LHL=2+3=5[/tex]
Right hand limit:
[tex]RHL=lim_{x\rightarrow 2^+}f(x)[/tex]
[tex]RHL=lim_{x\rightarrow 2^+}3-x[/tex]
Apply limit.
[tex]RHL=3-2=1[/tex]
[tex]5\neq 1[/tex]
Since LHL≠RHL, therefore the limit does not exist. Option 1 is correct.
