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I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. It's possible to resolve this limit with the developments of McLaurin? Can you explain the method and the steps used? Thanks

$$\lim\limits_{x \to 0} \left(\frac{(\sin x)^2-x^2}{x^4}\right)$$

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  • $\begingroup$ See math.stackexchange.com/questions/387333/… $\endgroup$ – lab bhattacharjee Dec 30 '15 at 17:05
  • $\begingroup$ Since $\sin^{2}(x)$ converges everywhere to its series expansion (i.e., analytic) you know that you can kill the $-x^2$ in the numerator via the expression for the series of $\sin^{2}(x)$. $\endgroup$ – Nobody Dec 30 '15 at 17:12
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You could very well use the series expansion of the function at $0$. Note that $$ \sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\dots $$ so $$ \sin^2(x)=x^2-\frac{x^4}{3}+\text{ higher order terms} $$ so $$ \sin^2(x)-x^2=-\frac{x^4}{3}+\text{ higher order terms} $$ so $$ \frac{\sin^2(x)-x^2}{x^4}=-\frac{1}{3}+\text{ higher order terms} $$ so $$ \lim_{x\to0}\frac{\sin^2(x)-x^2}{x^4}=? $$

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  • $\begingroup$ fixed, thanks for the correction $\endgroup$ – TomGrubb Dec 30 '15 at 17:14
  • $\begingroup$ I dont understand the second step: why is $x^2 -\frac{x^4}{3}$ $\endgroup$ – user12 Dec 30 '15 at 17:15
  • $\begingroup$ @Amarildo Square the series expression for $\sin(x)$. You'll get the $x^2$ term by multiplying $x$ and $x$. You'll get two $x^4$ terms by multiplying $x$ and $-x^3/3!$, which gives you the term $2\cdot x\cdot -x^3/3!=-x^4/3$. $\endgroup$ – kccu Dec 30 '15 at 17:17
  • $\begingroup$ the result with $x^6/36$ delete it right because its too big? $\endgroup$ – user12 Dec 30 '15 at 17:23
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    $\begingroup$ Pedagogically, the series actually shows what's going on though. Near zero, the series tells you the behavior of sin, which lets you deduce the behavior of the function above. L'Hopital hides this. Also, every step above was either squaring a series or simple algebraic maneuvers; it doesn't get much more simple than that $\endgroup$ – TomGrubb Dec 30 '15 at 17:37
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$$\lim_{x\to 0}\frac{(\sin x)^2-x^2}{x^4}$$ $$=\lim_{x\to 0}\frac{\frac{1-\cos 2x}{2}-x^2}{x^4}$$ $$=\frac 12\lim_{x\to 0}\frac{1-\cos 2x-2x^2}{x^4}$$ Using L'Hospital's rule for $\frac{0}{0}$ form four times, one should get
$$=\frac 12\lim_{x\to 0}\frac{-2^4\cos 2x}{4\cdot 3\cdot 2\cdot 1}$$

$$=-\frac{16\cos 0}{48}=\color{red}{-\frac 13}$$

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  • $\begingroup$ $16/48\neq 1/2$ $\endgroup$ – TomGrubb Dec 30 '15 at 17:18
  • $\begingroup$ @bburGsamohT: thanks for observation $\endgroup$ – Harish Chandra Rajpoot Dec 30 '15 at 17:19
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You can get there very easily by multiple application of the $\frac{0}{0}$ L'Hopital's rule:

$$\lim_{x \to 0} \frac{\sin^2 x - x^2}{x^4} = \lim_{x \to 0} \frac{\sin(2x) - 2x}{4x^3} = \lim_{x \to 0} \frac{2\cos(2x) - 2}{12x^2} = \lim_{x \to 0} \frac{-4\sin(2x)}{24x} = \lim_{x \to 0} \frac{-8\cos(2x)}{24} = -\frac{1}{3} $$

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  • $\begingroup$ The last use of L'Hospital's rule is not permitted; the form is not 0/0. $\endgroup$ – Bernard Masse Dec 30 '15 at 17:24
  • $\begingroup$ @BernardMasse No that was a notation mistake, i did not remove the -2 from the numerator $\endgroup$ – Adrian Dec 30 '15 at 17:32

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