Function limits: Solve the activities in each image (full development)
2 answers:
9514 1404 393
Answer:
1a. f(4) = 1/2
1b. DNE
2a. f(2) = 5
2b. f(-3/2) = 9 1/4
2c. 4
2d. -2
2e. -1/3
Step-by-step explanation:
The limit for functions that can be evaluated is simply the value of the function. (1a, 2a, 2b are like that)
Some rational functions, particularly the ratios of polynomials, can be simplified by cancelling common factors from numerator and denominator. The simplified form can often be evaluated directly at the variable value that leaves the original function "undefined." (2c and 2d are like that)
Where a (simplified) rational function has an odd-degree zero in the denominator, there is no limit at that point. The function value changes sign at the discontinuity, so the left limit is different from the right limit. The limit Does Not Exist. (1b is like that)
Indeterminate forms (0/0 or ∞/∞)have their own methods for finding limits. Chief among these is L'Hopital's rule, which has you compare derivatives of numerator and denominator, repeating as necessary if those also give an indeterminate form. I like to gain a clue from the graph. Often, the function can be evaluated "arbitrarily close" to the limiting value of the variable, so the limit can be guessed at without much trouble.
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1a. Evaluate the function: 1/(4-2) = 1/2
1b. the denominator changes sign at x=2, so the limit Does Not Exist
2a. Evaluate the function: (2-2)(2) +5 = 5
2b. Evaluate the function: (-3/2 -4)(-3/2) +1 = 33/4 +1 = 37/4 = 9 1/4
2c. Simplify and evaluate: (x -3)(x -7)/(x -7) = x -3 ⇒ 7 -3 = 4
2d. Simplify and evaluate: (x +1)(x +3)/(x +3) = x +1 ⇒ -3 +1 = -2
2e. The function has a hole at x=4 but can be evaluated nearby (approximation). (See attached). The limit is -1/3. (The ratio of derivatives per L'Hopital's rule reduces to -(√5-x)/√(5+x) = -√(1/9) = -1/3.)
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