Q. 37

Question

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

$\lim_{x \to 0} x^{- 3} .$

Step-by-Step Solution

Verified

Answer

Limit does not exist.

1Step 1. Given Information.

$G i v e n : \lim_{x \to 0} x^{- 3} .$

2Step 2. Theorem 1.16

$P o w e r f u n c t i o n s a r e c o n t i n u o u s o n t h e i r d o m a i n s . I n t e r m s o f l i m i t s, i f A i s r e a l a n d k i s r a t i o n a l, t h e n f o r a l l v a l u e s x = c a t w h i c h x^{k} i s d e f i n e d w e h a v e \lim_{x \to c} A x^{k} = A c^{k} .$

3Step 3. Finding limit of given function.

According to the theorem power functions are continuous in their domains but $x = 0$ is not in the domain of the function $x^{- 3}$ . Therefore, theorem does not apply for this limit and also limit does not exist for this function as it is not even in the domain.

Q. 36

Q. 38

Other exercises in this chapter

Q. 35

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.limx→-5 (3x-2)

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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. 38

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.limx→-5 x .

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Q. 39

In Exercises 39–44, use Theorem 1.16 and left and right limits to determine whether each function f is continuous at its break point(s). For each disconti

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