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$$\lim_{x\to+\infty} \dfrac{x - \ln(x+1)}{x \ln(x+1)}$$

I need help with this one .I couldn't figure out anything that could solve it .

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    – gebruiker
    Mar 21, 2016 at 18:47

2 Answers 2

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Note that $$\dfrac{x-\ln{(x+1)}}{x\ln{(x+1)}}=\dfrac{x}{x\ln{(x+1)}}-\dfrac{\ln{(x+1)}}{x\ln{(x+1)}}$$

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Note that $x \pm \ln (x + 1) \sim x$ (the logarithm is negligible) Then

$$\frac{x - \ln(x+1)}{x \ln(x+1)} \sim \frac{x}{x \ln(x+1)} = \frac{1}{\ln(x+1)}$$

So

$$\lim_{x \to \infty} \frac{x - \ln(x+1)}{x \ln(x+1)} = \lim_{x \to \infty} \frac{1}{\ln(x+1)} = 0$$

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