February 17th, 2017, 01:50 AM  #1 
Newbie Joined: Jan 2017 From: India Posts: 12 Thanks: 1  Limit problem!
What is the value of (Sinx/x)^1/(x^2) Where x tends to 0? 
February 17th, 2017, 03:40 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2197 Math Focus: Mainly analysis and algebra 
Is that $\displaystyle \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}}$? As a first thought, have you tried a series expansion of $\sin x$ or l'Hôpital's rule? Last edited by v8archie; February 17th, 2017 at 03:50 AM. 
February 17th, 2017, 02:43 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,236 Thanks: 498 
Try taking log. I get limit = $\displaystyle e^{\frac{1}{6}}$.

February 17th, 2017, 02:57 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2197 Math Focus: Mainly analysis and algebra 
That's what the power series $\displaystyle \sin x = x\tfrac{1}{3!}x^3 + \mathcal{O}(x^5)$ would suggest  and rather more easily than l'Hôpital's rule. (I did it in my head.)
Last edited by v8archie; February 17th, 2017 at 02:59 PM. 
February 17th, 2017, 06:25 PM  #5 
Member Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 
Can you elaborate on how you did the power series in your head to come up with $\displaystyle e^{\frac{1}{6}}$, please? How did you do the power series of it when it's being raised to by $\displaystyle \frac{1}{x^2}$? Thank you 
February 17th, 2017, 07:22 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2197 Math Focus: Mainly analysis and algebra 
There is a standard result that you should learn and remember (it is a definition of $e$): $$\lim_{x \to \infty} \left(1 + \frac{1}{x} \right)^{x} = e$$ If we then set $y = \frac{1}{x}$ we get $y \to 0$ as $x \to \infty$ (there's a small shortcut in that statement, but don't worry about it) and then $$\lim_{y \to 0} \left(1 + y\right)^{\frac{1}{y}} = e$$ The power series is something you might remember (or look up). It gives you \begin{align*} \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}} &= \lim_{x \to 0} \left( \frac{x  \frac{1}{3!}x^3 + \mathcal{O}(x^5)}{x} \right)^{\frac{1}{x^2}} \\ &= \lim_{x \to 0} \left( 1  \tfrac{1}{3!}x^2 + \mathcal{O}(x^4) \right)^{\frac{1}{\frac{1}{3!}x^2 + \mathcal{O}(x^4)} \cdot \frac{\frac{1}{3!}x^2 + \mathcal{O}(x^4)}{x^2}} \\ &= \lim_{x \to 0} \left( \left( 1  \tfrac{1}{3!}x^2 + \mathcal{O}(x^4) \right)^{\frac{1}{\frac{1}{3!}x^2 + \mathcal{O}(x^4)}} \right)^{\frac{1}{3!} + \mathcal{O}(x^2)} \\ &= \left( e \right)^{\frac{1}{6} + 0} \\ &= e^{\frac16} \end{align*} 
February 17th, 2017, 09:17 PM  #7 
Newbie Joined: Jan 2017 From: India Posts: 12 Thanks: 1 
@V8archie Thanks! I got it! 
February 18th, 2017, 02:36 PM  #8 
Global Moderator Joined: May 2007 Posts: 6,236 Thanks: 498 
Using log. $\displaystyle f(x)=(\frac{\sin x}{x})^{\frac{1}{x^2}}$ $\displaystyle \ln(f(x))=\frac{1}{x^2}\ln(\frac{\sin x}{x})$ $\displaystyle \ln(\frac{\sin x}{x})\approx \ln(1\frac{x^2}{6})\approx \frac{x^2}{6}$ Therefore: $\displaystyle \ln(f(x))\approx \frac{1}{6}$ or $\displaystyle f(x)\approx e^{\frac{1}{6}}$ Last edited by skipjack; February 18th, 2017 at 09:00 PM. 

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