Q. 35

Question

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.

$\lim_{x \to - 5} (3 x - 2) .$

Step-by-Step Solution

Verified

Answer

The limit exists and it is equal to -17.

1Step 1. Given Information.

$G i v e n : \lim_{x \to - 5} (3 x - 2)$

2Step 2. Theorem 1.16

$C o n s \tan t, i d e n t i t y, a n d l i n e a r f u n c t i o n s a r e c o n t i n u o u s e v e r y w h e r e . I n t e r m s o f l i m i t s, f o r e v e r y k, c, m, a n d b i n R w e h a v e \lim_{x \to c} k = k, \lim_{x \to c} x = c a n d \lim_{x \to c} (m x + b) = (m c + b) .$

3Step 3. Finding limit of given function.

$U \sin g t h e a b o v e t h e o r e m' s 3 r d c a s e w e h a v e, c = - 5, m = 3 a n d b = - 2 . P u t t i n g t h e s e v a l u e s w e g e t, \lim_{x \to - 5} (3 x - 2) = 3 (- 5) - 2 = - 15 - 2 = - 17 .$

Q. 34

Q. 36

Other exercises in this chapter

Q. 33

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.limx→-1 6.

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Q. 34

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.limx→-1 x.

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Q. 36

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.limx→3 x4 .

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Q. 37

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.limx→0 x-3 .

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