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So I have a limit that I want to solve:

$$\lim_{x\to0}\bigg(\frac{1+\ln(1-x^2)}{2x+1-\sin(x)}\bigg)^{\frac{1}{x^2}}$$

So I thought of using L'Hospital's rule, but it's not the $\displaystyle\frac{0}{0}$ situation.

Or Can it go for a different L'Hospital's rule situation, like it's $\displaystyle\frac{\infty}{\infty}$ but i get $1$ at the numerator? Can I still use it or should I first manipulate the fraction somehow that it's plausible for using that rule, or is there another trick to use?

Any help would be appreciated.

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  • $\begingroup$ If you like to use L'Hospital's rule then take the logarithm of your term because you will get $0/0$. But it becomes complicate. $\endgroup$ – user90369 Aug 25 '16 at 11:27
  • $\begingroup$ I have done as I have said above. With the first step I get $-\frac{1}{0}$. Therefore perhaps your task is written wrong. $\endgroup$ – user90369 Aug 25 '16 at 11:41
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Hint it's of the form $1^{\infty}$ so write we can write it as $e^{g(x)(f(x)-1)}$ (which can be proved using basic results of limits ) where $f(x)=base ,G(x)=index)$ then you have a form where both numerator,denominator tend to $0$ and thus you can use Lhospital or Taylor expansions for $ln,sin$

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You may do an expansion in power series (use $\log(1+z)=z-O(z^2)$ and $\sin(x)=x+O(x^3)$):

$$ \frac{1+\log(1-x^2)}{2x+1-\sin x} = \frac{1-x^2+O(x^4)}{1+x+O(x^3)} =1-x+O(x^3)$$

Now, taking log of your expression we may write it as $$ \frac{\log(1-x+O(x^3))}{x^2}=\frac{-1}{x}+O(1)$$ which goes to $\mp \infty$ as $x\rightarrow \pm 0$. So the limit is zero or infinity depending on how $x$ goes to zero.

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First of all, notice that a two-sided limit does not exist.

  • Use Taylor series (at $x=0$): $$\left(\frac{1+\ln(1-x^2)}{2x+1-\sin(x)}\right)^{\frac{1}{x^2}}=\exp\left[-\frac{1}{x}-\frac{1}{2}-\frac{x}{2}-\frac{7x^2}{12}-\frac{43x^3}{120}-\frac{149x^4}{180}+O(x^5)\right]$$
  • Using l'Hôpital's rule: $$\lim_{x\to0^-}\left(\frac{1+\ln(1-x^2)}{2x+1-\sin(x)}\right)^{\frac{1}{x^2}}=\exp\left[\lim_{x\to0^-}\frac{\ln\left(\frac{1+\ln(1-x^2)}{2x+1-\sin(x)}\right)}{x^2}\right]=\exp\left[-\frac{\lim_{x\to0^-}\frac{1}{x}}{2}\right]\to\infty$$
  • Using l'Hôpital's rule: $$\lim_{x\to0^+}\left(\frac{1+\ln(1-x^2)}{2x+1-\sin(x)}\right)^{\frac{1}{x^2}}=\exp\left[\lim_{x\to0^+}\frac{\ln\left(\frac{1+\ln(1-x^2)}{2x+1-\sin(x)}\right)}{x^2}\right]=\exp\left[-\frac{\lim_{x\to0^+}\frac{1}{x}}{2}\right]=0$$
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As others have said, find the limit of the log of your expression. But take advantage of the log:

$$\lim_{x\rightarrow 0} \frac{\ln(1+\ln(1-x^2)) - \ln(2x+1-\sin x)}{x^2}.$$

Apply L'hospital once to get

$$\lim_{x\rightarrow 0} \frac{ \frac{1}{1+\ln(1-x^2)}\frac{1}{1-x^2}(-2x) - \frac{1}{2x+1-\sin x}(2-\cos x)}{2x}.$$

Then the numerator goes to $-1$ and the denominator goes to $0.$ So the left hand limit is $+\infty$ and the right hand limit is $-\infty$. So for your original expression, the left hand limit is $e^{+\infty} = +\infty$ and the right hand limit is $e^{-\infty} = 0$.

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