Use l’Hôpital’s rule to find the limits In (x? + 2x)limIn x

Use L’Hôpital’s rule to find the limits limx→0+lnx2+2xlnx.

Answered: In (x? + 2x) lim In x

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.1: Polynomial Functions Of Degree Greater Than

Problem 55E

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Question

Use l’Hôpital’s rule to find the limits

Expert Solution

Step 1

Use L’Hôpital’s rule to find the limits $\lim_{x \to 0^{+}} \frac{\ln (x^{2} + 2 x)}{\ln x}$ .

Step 2

$\begin{array}{rcl} \lim_{x \to 0^{+}} \frac{\ln (x^{2} + 2 x)}{\ln x} & = & \frac{\ln (0^{2} + 2 (0))}{\ln (0)} \\ = & \frac{0}{0} \end{array}$

Apply L'Hospital's Rule.

$\begin{array}{rcl} \lim_{x \to 0^{+}} \frac{\ln (x^{2} + 2 x)}{\ln x} & = & \lim_{x \to 0^{+}} \frac{\frac{d}{d x} (\ln (x^{2} + 2 x))}{\frac{d}{d x} (\ln x)} \\ = & \lim_{x \to 0^{+}} \frac{\frac{1}{x^{2} + 2 x} \frac{d}{d x} (x^{2} + 2 x)}{(\frac{1}{x})} \\ = & \lim_{x \to 0^{+}} \frac{\frac{1}{x^{2} + 2 x} (2 x + 2)}{(\frac{1}{x})} \\ = & \lim_{x \to 0^{+}} \frac{x (2 x + 2)}{x^{2} + 2 x} \\ = & \frac{0 (2 (0) + 2)}{{(0)}^{2} + 2 (0)} \\ = & \frac{0}{0} \end{array}$

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