2
$\begingroup$

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

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?

$\endgroup$
2
  • 2
    $\begingroup$ The limit is $2$. $\endgroup$
    – Did
    Jun 1, 2012 at 6:31
  • $\begingroup$ Yes, you are right, I have amended my post. $\endgroup$ Jun 2, 2012 at 1:50

4 Answers 4

6
$\begingroup$

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

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

$\endgroup$
5
  • $\begingroup$ Something when wrong in the TeX-ing $\endgroup$ Jun 1, 2012 at 6:42
  • 1
    $\begingroup$ @AD. I tried to make the answer hide but somehow I don't know why it doesn't work. $\endgroup$
    – user17762
    Jun 1, 2012 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. $\endgroup$ Jun 1, 2012 at 6:55
  • $\begingroup$ @MartinSleziak Thanks. I have divided it into separate lines. $\endgroup$
    – user17762
    Jun 1, 2012 at 6:58
  • $\begingroup$ @Gigili Thanks. I have done what Martin suggested. $\endgroup$
    – user17762
    Jun 1, 2012 at 6:58
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $2$.

$\endgroup$
0
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.