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Chapter 4: Problem 43

Evaluate the following limits. $$\lim _{x \rightarrow \infty} \frac{x^{2}-\ln (2 / x)}{3 x^{2}+2 x}$$

Short Answer

Expert verified
Question: Determine the limit of the function as x approaches infinity: $$\lim_{x \rightarrow \infty} \frac{x^{2}-\ln(2/x)}{3x^{2}+2x}$$ Answer: The limit of the function as x approaches infinity is \(\frac{1}{3}\).

Step by step solution

01

Identify the dominating terms

To identify the dominating terms, look at the numerator and denominator. As x approaches infinity, the x^2 and ln(2/x) terms in the numerator will be the most significant. Similarly, the 3x^2 term in the denominator will be the most significant. Numerator dominating term: \(x^2\) Denominator dominating term: \(3x^2\)
02

Divide the numerator and denominator by the dominating term of the denominator

To simplify the limit, divide both the numerator and the denominator by the dominating term of the denominator, which is \(3x^2\). $$\lim_{x \rightarrow \infty} \frac{\frac{x^{2}}{3x^{2}}-\frac{\ln(2/x)}{3x^{2}}}{\frac{3x^{2}}{3x^{2}}+\frac{2x}{3x^{2}}}$$
03

Simplify the expression

Now, simplify the expression as much as possible: $$\lim_{x \rightarrow \infty} \frac{\frac{1}{3}-\frac{\ln(2/x)}{3x^{2}}}{1+\frac{2}{3x}}$$
04

Evaluate the limit

As x approaches infinity, the \(\frac{\ln(2/x)}{3x^{2}}\) term will tend to 0, while the \(\frac{2}{3x}\) term in the denominator will also tend to 0. Thus, the limit can be more easily evaluated: $$\lim_{x \rightarrow \infty} \frac{\frac{1}{3}-0}{1+0} = \boxed{\frac{1}{3}}$$

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