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Chapter 1: Q. 36 (page 149)

Calculate each limit in Exercises 35-80.
$\lim_{x \to 0} (2 x^{- \frac{3}{4}})$

Short Answer

Expert verified

The function $\lim_{x \to 0} (2 x^{- \frac{3}{4}})$ does not exist.

Step by step solution

01

Step 1. Given information.

We have:

$\lim_{x \to 0} (2 x^{- \frac{3}{4}})$

02

Step 2. Find the limits.

The function is $\lim_{x \to 0} (2 x^{- \frac{3}{4}})$ .

$If \lim_{x \to a -} f (x) \neq \lim_{x \to a +} f (x) then the limit does not exist$

$\lim_{x \to 0^{+}} (2 x^{- \frac{3}{4}}) = 2 \lim_{x \to 0} (\frac{1}{x^{\frac{3}{4}}}) = \infty \lim_{x \to 0^{-}} (2 x^{- \frac{3}{4}}) = 2 \lim_{x \to 0^{-}} (\frac{1}{x^{\frac{3}{4}}}) = Does not exist$

Therefore, the limit does not exist.

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Most popular questions from this chapter

Use the delta-epsilon definition of continuity to argue that f is or is not continuous at the indicated point $x = c$ .

$f (x) = \{\begin{aligned} 2 - x, if x rational \\ x^{2}, if x irrational, \end{aligned} c = 2$

For each limit $\underset{x \to c}{l i m} f (x) = L$ in Exercises 43–54, use graphs and algebra to approximate the largest value of $δ$ such that if $x \in (c - δ, c) \cup (c, c + δ) then f (x) \in (L - ε, L + ε)$

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$\lim_{x \to 4} (3 e^{1.7 x} + 1)$

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

$\lim_{x \to - 1} x .$

For each limit statement , use algebra to find δ > 0 in terms of $ε$ > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < $ε$ .

$\lim_{x \to 3} (x^{2} - 2 x - 3) = 0; you may assume δ \leq 1$

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