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I tried :$$\lim_{x\to 0^+}\frac{\ln x}{(1+x)^2}-\ln\left(\frac{x}{1+x}\right)=\lim_{x\to 0^+}\frac{\ln x}{(1+x)^2}-\ln x+\ln (x+1)=\lim_{x\to 0^+}\frac{-x^2\ln x}{(1+x)^2}+\ln(x+1)$$

Here I can use $\ln(x+1)\sim x-\frac{x^2}{2}+\frac{x^3}{3}$. but it seems It doesn't work because I don't know how to simplify $\cfrac{-x^2\ln x}{(1+x)^2}$.

User
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2 Answers2

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Actually, $(1+x)^2-1$ isn't $x^2$. I should be$$\lim_{x\to0^+}\left(\frac{-x(x+2)\ln x}{(1+x)^2}+\ln(x+1)\right)=-2\lim_{x\to0^+}x\ln x=0.$$(See here for proofs that $\lim_{x\to0^+}x\ln x=0$.)

J.G.
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  • I don't understand how you get $-2\lim_{x\to0^+}x\ln x$. Did you express that as$\cfrac{-x(x+2)\ln x}{(1+x)^2}=\cfrac{-\ln x\times x^2-2\ln x\times x}{x^2+2x+1}$? – User Nov 14 '20 at 17:13
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    @soheil I used $\lim_{x\to0^+}\tfrac{x+2}{(1+x)^2}=2$. – J.G. Nov 14 '20 at 17:18
  • I got it, Can we always say limit of product of two functions is equal to product of their limit? – User Nov 14 '20 at 17:23
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    @soheil You can say $\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\lim_{x\to a}g(x)$ provided $F:=\lim_{x\to a}f(x)$ and $G:=\lim_{x\to a}g(x)$ exist and their product isn't the indeterminate form $0\times\pm\infty$. In this case $F=2,,G=0$. For the case where $F,,G$ are finite, see the product rule of limits here. – J.G. Nov 14 '20 at 17:30
  • Very good explanation, Thank you! – User Nov 14 '20 at 20:06
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Answer is $0$

And the last step should give you

$\ln(x)-\ln(\frac{x}{1+x})(1+x^2)$ and as $x=0$ it gives you $0$!

User
  • 8,033