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I'm looking for a solution to the following limit from above: \begin{equation} \lim_{x\rightarrow1}\frac{\frac{4x^3-2x}{\sqrt{x^2-1}}\ln(x+\sqrt{x^2-1})-2x^2}{x^4-1} \end{equation}

By plotting it, I know it approaches $4/3$.

L'hopital becomes difficult quickly. Does anyone see how to approach at this answer analytically? Any simplifications or approaches (or possibly approximations) are appreciated!

Thanks in advance!

Blue
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user2704177
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    Perhaps use the fact that if $y=\ln(x+\sqrt{x^2-1}))$, then $x=\cosh(y)$. Substitute $\cosh y$ for $x$ and let $y\to0$. – John Wayland Bales Jan 24 '22 at 16:27
  • Or perhaps not, still looks like a mess. – John Wayland Bales Jan 24 '22 at 16:39
  • Replace denominator by $4(x-1)$ and use the substitution mentioned by @JohnWaylandBales. After some simplification you will get to $\frac{y\cosh 2y-\cosh y\sinh y} {y^3}$ and this should be easier to handle. Let us know (by updating your post) if you face any further issue. – Paramanand Singh Jan 25 '22 at 06:56

1 Answers1

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To make things a little bit easier, first factor $x$ (whose limit is $1$) out of the numerator, then use $x^4-1=(x-1)(x+1)(x^2+1)$ for the denominator: \begin{align} \lim_{x\to 1}\frac{\frac{4x^3-2x}{\sqrt{x^2-1}}\ln(x+\sqrt{x^2-1})-2x^2}{x^4-1}&=\lim_{x\to 1}\frac{\frac{4x^2-2}{\sqrt{x^2-1}}\ln(x+\sqrt{x^2-1})-2x}{x^4-1}\\ &=\frac{1}{4}\lim_{x\to 1}\frac{\frac{4x^2-2}{\sqrt{x^2-1}}\ln(x+\sqrt{x^2-1})-2x}{x-1} \end{align} Denote $t=x+\sqrt{x^2-1}$.

Using $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$ we can write $$\ln(x+\sqrt{x^2-1})=\ln(1+(t-1))=(t-1)-\frac{(t-1)^2}{2}+\frac{(t-1)^3}{3}+o\left((t-1)^3\right).$$ Notice that $o\left((t-1)^3\right)=o\left((x-1)^{3/2}\right)$ so that substituting this back into the limit we get $$\lim_{x\to 1}\frac{\frac{4x^2-2}{\sqrt{x^2-1}}\left(x+\sqrt{x^2-1}-1-\frac{(x+\sqrt{x^2-1}-1)^2}{2}+\frac{(x+\sqrt{x^2-1}-1)^3}{3}+o\left((x-1)^{3/2}\right)\right)-2x}{x-1}$$ where I put the factor $\frac{1}{4}$ aside for now. We have $$\lim_{x\to 1} \frac{\frac{4x^2-2}{\sqrt{x^2-1}}o\left((x-1)^{3/2}\right)}{x-1}=0$$ and this is why we had to take $3$ terms from the Taylor series. So now we can consider just $$\lim_{x\to 1}\frac{\frac{4x^2-2}{\sqrt{x^2-1}}\left(x+\sqrt{x^2-1}-1-\frac{(x+\sqrt{x^2-1}-1)^2}{2}+\frac{(x+\sqrt{x^2-1}-1)^3}{3}\right)-2x}{x-1}$$ which can be solved using standard techniques: simplifying we get $$\frac{2}{3}\lim_{x\to 1} \frac{8x^4-10x^3-2x^2+5x-1+2\sqrt{x^2-1}(4x^3-5x^2+x+4)}{\sqrt{x^2-1}}$$ which is the same as $$\frac{\sqrt{2}}{3}\lim_{x\to 1} \frac{8x^4-10x^3-2x^2+5x-1+2\sqrt{x^2-1}(4x^3-5x^2+x+4)}{\sqrt{x-1}}$$ putting $\frac{\sqrt{2}}{3}$ aside and using L'Hopital's rule we get $$\lim_{x\to 1} 2\sqrt{x-1}\left(32x^3-30x^2-4x+5+2\sqrt{x^2-1}(12x^2-10x+1)\\ +\frac{2x(4x^3-5x^2+x+4)}{\sqrt{x^2-1}}\right)$$ everything drops and we are left just with $$\lim_{x\to 1} \frac{4x\sqrt{x-1}(4x^3-5x^2+x+4)}{\sqrt{x^2-1}}=\lim_{x\to 1} \frac{4x(4x^3-5x^2+x+4)}{\sqrt{x+1}}=\frac{16}{\sqrt{2}}$$

Bringing back the constants we took out, the original limit equals $\frac{1}{4}\cdot\frac{\sqrt{2}}{3}\cdot\frac{16}{\sqrt{2}}=\frac{4}{3}$.

Snaw
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