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MCD2130 - T2 - 2025 Lecture Task Week 5

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Question text Consider the function [math: f(x)={3x+5,x<33x2+4x−2,x≥3] f(x)= \begin{cases} \displaystyle & {3\,x+5}, & x < {3} \\ & {3\,x^2+4\,x-2}, & x \geq {3}\end{cases} . a) [math: limx→3−f(x)=]\displaystyle \lim_{{x \to {3}^-}} f(x) = [input] b) [math: limx→3+f(x)=]\displaystyle\lim_{{x \to {3}^+}} f(x) = [input] c) [math: limx→3f(x)=]\displaystyle\lim_{{x \to {3}}} f(x) = [select: (Clear my choice), does not exist since left limit is not equal to right limit., exists and equals 37] Check Question 3

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First, I restate the problem to ensure understanding: we have a piecewise function f(x) defined by 3x+5 for x < 3 and 3x^2+4x-2 for x ≥ 3. We are asked for three limits at x = 3: (a) the left-hand limit, (b) the right-hand limit, and (c) the overall limit. Option a) lim_{x→3^-} f(x) = 14. To evaluate the left-hand limit, we use the branch applicable for x < 3, namely f(x) = 3x +......Login to view full explanation

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Which of the following statements must be true, and which are false? Note: each part is independent from the others. a) If 𝑓 ( 𝑥 ) is a polynomial, then lim 𝑥 → 5 𝑓 ( 𝑥 )  can be evaluated by computing 𝑓 ( 5 ) . [ Select ] False True b) If 𝑅 ( 𝑥 )   is a rational function, then lim 𝑥 → 7 𝑅 ( 𝑥 )     can be evaluated by computing 𝑅 ( 7 )  . [ Select ] False True c) If 𝑝 ( 𝑥 )  is a polynomial, then lim 𝑥 → 5 + 𝑝 ( 𝑥 )    can be evaluated by computing 𝑝 ( 5 )  . [ Select ] True False d) If 𝑓 ( 2 ) = 4  then lim 𝑥 → 2 [ 𝑓 ( 𝑥 ) ] 2 = 16   [ Select ] False True e) If   lim 𝑥 → 4 ( 5 𝑓 ( 𝑥 ) ) = 15   , then lim 𝑥 → 4 𝑓 ( 𝑥 ) = 3   . [ Select ] True False f) If   lim 𝑥 → 4 ( 𝑥 𝑓 ( 𝑥 ) ) = 8   , then lim 𝑥 → 4 𝑓 ( 𝑥 ) = 2    . [ Select ] False True

  the graph of f(x)f\left(x\right) given above. Use it to find the following one-sided and two-sided limits. (If a limit does not exist, write DNE.) limx→1−f(x)=\lim\limits_{x\:\rightarrow1^-}\:f\left(x\right)= [Fill in the blank], limx→1+f(x)=\lim\limits_{x\rightarrow1^+}f\left(x\right)= [Fill in the blank], limx→1f(x)=\lim\limits_{x\rightarrow1}f\left(x\right)\:= [Fill in the blank], limx→2f(x)=\lim\limits_{x\rightarrow2}f\left(x\right)= [Fill in the blank], limx→3−f(x)=\lim\limits_{x\rightarrow3^-}f\left(x\right)= [Fill in the blank], limx→3f(x)=\lim\limits_{x\rightarrow3}f\left(x\right)=[Fill in the blank], limx→4f(x)=\lim\limits_{x\rightarrow4}f\left(x\right)=[Fill in the blank],

                  Consider the two graphs above. What are the following limits? (If a limit does not exist, write DNE.) limx→1f(x)=\lim\limits_{x\rightarrow1}f\left(x\right)= [Fill in the blank], limx→1g(x)=\lim\limits_{x\rightarrow1}g\left(x\right)= [Fill in the blank], Note that the two functions f(x)f\left(x\right) and g(x)g\left(x\right) are identical except for at x=1x=1    . Is the following statement TRUE or FALSE? For any function  h(x)h\left(x\right) , the limit limx→ah(x)\lim\limits_{x\rightarrow a}h\left(x\right) does not depend on the value of h(x)h\left(x\right) at  x=ax=a  , or even whether h(a)h\left(a\right) is defined or not.  [Fill in the blank], (Write "TRUE" or "FALSE".)  

Question text Consider the graph of [math: f(x)] shown below: a) [math: limx→2−f(x)=]\displaystyle \lim_{{x \to {2}^-}} f(x) =[input] b) [math: limx→2+f(x)=]\displaystyle\lim_{{x \to {2}^+}} f(x) =[input] c) [math: limx→2f(x)=]\displaystyle\lim_{{x \to {2}}} f(x) =[select: (Clear my choice), does not exist., exists and equals 1] Check Question 1

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