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Chapter 1: Q. 55 (page 149)

Calculate each limit in Exercises 35鈥�80.
$\underset{x 鈫� 0^{+}}{l i m} \frac{x^{\frac{7}{2}} - x^{\frac{8}{3}}}{x^{2}}$

Short Answer

Expert verified

The limit is 0

Step by step solution

01

Step 1. Given information

The given expression is $\underset{x 鈫� 0^{+}}{l i m} \frac{x^{\frac{7}{2}} - x^{\frac{8}{3}}}{x^{2}}$

02

Step 2. Calculation

The limit is calculated as below by using the L'Hopital rule,

$\underset{x 鈫� 0^{+}}{l i m} \frac{x^{\frac{7}{2}} - x^{\frac{8}{3}}}{x^{2}} = \underset{x 鈫� 0^{+}}{l i m} (\frac{\frac{7}{2} x^{\frac{5}{2}} - \frac{8}{3} x^{\frac{5}{3}}}{2 x}) (differentiating the numerator and denominator) = \underset{x 鈫� 0^{+}}{l i m} (\frac{\frac{35}{4} x^{\frac{3}{2}} - \frac{45}{9} x^{\frac{2}{3}}}{2}) (differentiating the numerator and denominator) = (\frac{\frac{35}{4} {(0)}^{\frac{3}{2}} - \frac{45}{9} {(0)}^{\frac{2}{3}}}{2}) = \frac{0 - 0}{2} = 0$

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Most popular questions from this chapter

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

$\lim_{x 鈫� 0} (3 - 4 x) = 3$

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

$\lim_{x 鈫� - 5} (3 x - 2) .$

For each limit $\underset{x 鈫� c}{l i m} f (x) = L$ in Exercises 43鈥�54, use graphs and algebra to approximate the largest value of $未$ such that if $x 鈭� (c - 未, c) 鈭� (c, c + 未) then f (x) 鈭� (L - 蔚, L + 蔚)$

$\underset{x 鈫� - 1}{l i m} \frac{x^{2} - 2 x - 3}{x + 1} = - 4, 蔚 = 0.1$

Calculate each of the limits:

$\lim_{x 鈫� \frac{蟺}{2}} \frac{c o t x}{\cos x}$ .

Write delta-epsilon proofs for each of the limit statements in Exercises $47 鈥� 60 .$

$\lim_{x 鈫� 5^{+}} \sqrt{x - 5} = 0$

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