Calculate the following limit : $$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$$

This is what I have tried: Using Maclaurin series for $ (1+x)^a $: $$(1+x^2)^{1/2}=1+\frac{1}{2!}x^2\quad \text{(We'll stop at order 2)}$$ Using Maclaurin series for $\cos x $: $$\cos x=1-\frac{x^2}{2!}\quad \text{(We'll stop at order 2)}$$ This leads to : $$1-\sqrt{1+x^2}\cos x=1-(1+\frac{x^2}{2})(1-\frac{x^2}{2})=\frac{x^4}{4}$$ $$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4$$ Using Maclaurin series for $\sin x $: $$\sin x=x-\frac{x^3}{3!}\quad \text{(We'll stop at order 3)}$$ $$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4 = \left(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}}\right)^4$$

Thus $$\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\frac{\frac{x^4}{4}}{(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}})^4}=\frac{x^4(1-\frac{x^2}{2})^4}{4(x-\frac{x^3}{3!})}=\frac{(x(1-\frac{x^2}{2}))^4}{4(x-\frac{x^3}{3!})}=\frac{1}{4}(\frac{x-\frac{x^3}{2}}{x-\frac{x^3}{3!}})=\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}}) $$

Then $$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\lim_{x\rightarrow 0}\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})=\frac{1}{4}.$$ But my book says the solution is $\frac{1}{3}$

Where have I done wrong?

Help appreciated!


Hint. You need a longer expansion of $\sqrt{1+x^2}$ and $\cos(x)$: $$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4)\quad,\quad \cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4).$$ Then \begin{align}\sqrt{1+x^2}\cos(x)&=\left(1+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4)\right)\left(1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)\right)\\ &=\left(1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)\right)+\frac{x^2}{2}\left(1-\frac{x^2}{2}+o(x^2)\right)-\frac{x^4}{8}\left(1+o(1)\right)+o(x^4)\\ &=1-\frac{x^2}{2}+\frac{x^4}{4!}+\frac{x^2}{2}-\frac{x^4}{4}-\frac{x^4}{8}+o(x^4)\\ &=1-\frac{x^4}{3}+o(x^4). \end{align} Can you take it from here?

  • $\begingroup$ How can i know at what order should i stop the expansion at then ? I always seem to stop at the wrong order. $\endgroup$ – Raku Nov 15 '17 at 11:21
  • $\begingroup$ Use the little-oh notation: en.wikipedia.org/wiki/Big_O_notation#Little-o_notation $\endgroup$ – Robert Z Nov 15 '17 at 11:22
  • $\begingroup$ I meant how did you know you should expand till o(x4), why didn't you continue the expansion or stop before ? $\endgroup$ – Raku Nov 15 '17 at 11:24
  • $\begingroup$ The denominator says that we need a $4$th order precision also at the numerator. $\endgroup$ – Robert Z Nov 15 '17 at 11:27
  • $\begingroup$ So the order of the denominator determines the order of the numerator? $\endgroup$ – Raku Nov 15 '17 at 11:28

Without using L'Hospital & Taylor's Expansion,

$$=\dfrac{1-(1+x^2)(1-\sin^2x)}{1+\cos x\sqrt{1+x^2}}\cdot\dfrac{\cos^4x}{\sin^4x}$$

$$=\dfrac{x^2\sin^2x+(\sin x-x)(\sin x+x)}{x^4}\cdot\dfrac{\cos^4x}{\left(\dfrac{\sin x}x\right)^4(1+\cos x\sqrt{1+x^2})}$$

Now $\dfrac{(\sin x-x)(\sin x+x)}{x^4}=\left(\dfrac{\sin x}x+1\right)\cdot\dfrac{(\sin x-x)}{x^3}$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion


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