$$\lim\limits_{x\to\infty} \frac{\ln(x^2+4)}{\sinh^{-1}x}$$

This is an exam practice question. BTW, I am refering to the inverse hyperbolic function above.

Since this is infinity/infinity, I used one application of L'Hopital's rule for this one, and then did a bit of algebraic manipulation to get $\frac{2}{1}$, i.e. $2$. Is this right?

Is there a quicker way to do this? It seems like Taylor Polynomials are often used, but they confuse me - Is this one, where Taylor Polynomials would have been easier?

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    $\begingroup$ The limit is $2$. $\endgroup$ – Did Jun 1 '12 at 6:31
  • $\begingroup$ Yes, you are right, I have amended my post. $\endgroup$ – JackReacher Jun 2 '12 at 1:50

L'Hospital's Rule can be efficient. In this case, it helps a lot if the derivative of $\sinh^{-1} x$ is part of your standard toolkit. Recall that if $v=\sinh^{-1}x$, then $$\frac{dv}{dx}=\frac{1}{\sqrt{x^2+1}}.$$ Using the Chain Rule, we can see that if $u=\ln(x^2+4)$ then $$\frac{du}{dx}=\frac{2x}{x^2+4}.$$ Thus, using L'Hospital's Rule, we can see that our limit is $$\lim_{x\to\infty} \frac{2x\sqrt{x^2+1}}{x^2+4}.$$ Now the limit is probably obvious. But let us recall the standard formal technique for dealing with such things. Divide "top" and "bottom" by $x^2$. So we want $$\lim_{x\to\infty} 2\frac{\sqrt{1+\frac{1}{x^2}}}{1+\frac{4}{x^2}},$$ and this limit is clearly $2$.


HINT Use $\sinh^{-1}(x) = \ln (x + \sqrt{1+x^2})$.

$$\displaystyle \lim_{x \rightarrow \infty} \dfrac{\ln(x^2+4)}{\ln(x+\sqrt{1+x^2})} = \displaystyle \lim_{x \rightarrow \infty} \dfrac{\ln(x^2) + \ln(1+4/x^2)}{\ln(x) + \ln(1+\sqrt{1+1/x^2})}$$ $$ = \displaystyle \lim_{x \rightarrow \infty} \dfrac{\ln(x^2)}{\ln(x)} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)}$$ $$ = \displaystyle \lim_{x \rightarrow \infty} 2 \dfrac{\ln(x)}{\ln(x)} \lim_{x \rightarrow \infty} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)}$$ $$ = 2 \displaystyle \lim_{x \rightarrow \infty} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)} =2 $$

  • $\begingroup$ Something when wrong in the TeX-ing $\endgroup$ – AD. Jun 1 '12 at 6:42
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    $\begingroup$ @AD. I tried to make the answer hide but somehow I don't know why it doesn't work. $\endgroup$ – user17762 Jun 1 '12 at 6:47
  • $\begingroup$ @Marvis: I don't know the reason for this behavior, but it seems to work ok if I divide it into separate lines and put each of them as a spoiler, see here. $\endgroup$ – Martin Sleziak Jun 1 '12 at 6:55
  • $\begingroup$ @MartinSleziak Thanks. I have divided it into separate lines. $\endgroup$ – user17762 Jun 1 '12 at 6:58
  • $\begingroup$ @Gigili Thanks. I have done what Martin suggested. $\endgroup$ – user17762 Jun 1 '12 at 6:58

Another possibility: substitution $\sinh^{-1}(x)=t$.

You have:

  • $t\to\infty$ for $x\to\infty$
  • $x=\sinh t= \frac{e^t-e^{-t}}2$
  • $x^2+4=\frac{e^{2t}+e^{-2t}}4+3$

So your limit becomes $$\lim\limits_{t\to\infty} \frac{\ln\left(\frac{e^{2t}+e^{-2t}+12}4\right)}t= \lim\limits_{t\to\infty} \frac{\ln(e^{2t}+e^{-2t}+12)-\ln 4}t,$$ which should not be that difficult.


For the next solution we'll resort to sinh(x) that goes to $\infty$ when x tends to $\infty$. Just throw sinh(x) in the limit and then consider the dominant term on nominator. Then you get that:

$$\lim\limits_{x\to\infty} \frac{\ln(x^2+4)}{\sinh^{-1}x} = \lim\limits_{x\to\infty} \frac{\ln(\sinh^{2}x+4)}{x} = \lim\limits_{x\to\infty} \frac{\ln{e^{2x}}}{x} = \lim\limits_{x\to\infty} \frac{{2x}}{x}=2$$

The proof is complete.


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