Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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?

share|improve this question
2  
The limit is $2$. –  Did Jun 1 '12 at 6:31
    
Yes, you are right, I have amended my post. –  JackReacher Jun 2 '12 at 1:50
add comment

4 Answers

up vote 2 down vote accepted

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 $2$.

share|improve this answer
    
Thank you Andre –  JackReacher Jun 2 '12 at 1:51
add comment

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.

share|improve this answer
add comment

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

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

share|improve this answer
    
Something when wrong in the TeX-ing –  AD. Jun 1 '12 at 6:42
1  
@AD. I tried to make the answer hide but somehow I don't know why it doesn't work. –  user17762 Jun 1 '12 at 6:47
    
@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. –  Martin Sleziak Jun 1 '12 at 6:55
    
@MartinSleziak Thanks. I have divided it into separate lines. –  user17762 Jun 1 '12 at 6:58
    
@Gigili Thanks. I have done what Martin suggested. –  user17762 Jun 1 '12 at 6:58
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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