Business > CASE STUDY > Calculus 1 task 1.docx Running head: CALCULUS 1 TASK 1: LIMITS AND CONTINUITY Calculus (All)
Calculus 1 task 1.docx Running head: CALCULUS 1 TASK 1: LIMITS AND CONTINUITY Calculus
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Calculus 1 task 1.docx Running head: CALCULUS 1 TASK 1: LIMITS AND CONTINUITY Calculus 1 Task 1: Limits and Continuity A. Given f ( x )= x −2 x2−3 x+2 , find lim x �... ��2 x → 1 x2−3 x+2 or explain why the limit does not exist. The limit of x approaching the number 1 does not exist because the left and right hand sides do not line up when substituting numbers in for x. When I tried substituting slightly lower numbers than 1 when approaching from the left hand side and slightly higher numbers than 1 when approaching from the right hand side, the numbers did not line up. This indicates that the limit does not exist. I also tried plugging the number 1 directly in for x in the function, I ended up with a zero in the denominator. This also indicates that the limit does not exist. The work that I have completed is shown below: Example 1. lim x −2 x →1 x2−3 x +2 (1)−2 x →1 (1)2−3 (1 )+ 2 Step 1- substitute the number one in for x −1 x → 1 (1 )2−3(1 )+2 Step 2- one minus two equals negative one 3. lim −1 x → 1 1 −3 +2 4. lim −1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. .. . . . . . . . . . . . . . . . . . .. . . . . [Show More]
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