Chapter 1: Q. 65 (page 149)

Calculate each limit in Exercises 35–80.
$\underset{x \to \infty}{l i m} (\ln x^{2} - \ln (2 x + 1))$

Short Answer

Expert verified

The limit is infinity

Step by step solution

Step 1. Given information

The given expression is $\underset{x \to \infty}{l i m} (\ln x^{2} - \ln (2 x + 1))$

Step 2. Calculation

The limit is calculated as below,

$\underset{x \to \infty}{l i m} (\ln x^{2} - \ln (2 x + 1)) = \underset{x \to \infty}{l i m} (\ln (\frac{x^{2}}{2 x + 1})) = \underset{x \to \infty}{l i m} (\ln (\frac{x^{2}}{x^{2} (\frac{2 x}{x^{2}} + \frac{1}{x^{2}})})) = \underset{x \to \infty}{l i m} (\ln (\frac{1}{\frac{2}{x} + \frac{1}{x^{2}}})) = (\ln (\frac{1}{\frac{2}{\infty} + \frac{1}{\infty^{2}}})) = \ln (\frac{1}{0 + 0}) = \ln \infty = \infty$

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Calculate each limit in Exercises 35–80.
$\underset{x \to \infty}{l i m} (\ln x^{2} - \ln (2 x + 1))$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

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Calculate each limit in Exercises 35–80.limx→∞lnx2-ln(2x+1)

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

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Calculate each limit in Exercises 35–80.
$\underset{x \to \infty}{l i m} (\ln x^{2} - \ln (2 x + 1))$