$\lim \frac{x^2-\sin^2{x}}{\tan(3x^4)}$ as $x$ goes to $0$

Question

Calculate $\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)}$

How does one calculate this limit?

Is it valid to say, since $\sin^2{x}$ is approximated by $x^2$ as $x \to 0$, we have:

$\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)} =\lim_{x \to 0} \frac{x^2-x^2}{\tan(3x^4)} =0$

Since $\tan(y)\approx y$ close to $y=0$, you need to develop the numerator to fourth order (at least). — Claude Leibovici, Jan 05 '16 at 06:56
See:http://math.stackexchange.com/questions/1594112/how-to-compute-lim-limits-x-to-0-left-frac-sin-x2-x2x4-right/1594166#1594166 — medicu, Jan 05 '16 at 20:54

Archis Welankar · Answer 1 · 2016-01-05T06:36:20.797

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Hint it is $0/0$ form so separetely differentiate wrt to $x$ numerator,denominator this is L'hospitals rule and then plug in 0 and get the required limit. If this doesnt work use Taylor series expansion for $\sin\theta,tan\theta$

edited Jan 05 '16 at 06:36

answered Jan 05 '16 at 06:27

Archis Welankar

15,910

Did you try using L'Hospital's rule on this problem? You'll still get $0/0$ (and again, and again once more). It requires four applications. – Jan 05 '16 at 06:30
Dont think so!.. – Archis Welankar Jan 05 '16 at 06:35
Well, the first derivative of the numerator is $2x -2 \sin x \cos x$, which is zero at zero. Then the second derivative is $2 + 2 \sin^2 x - 2 \cos^2 x$, which is also zero at zero. The third derivative is too. So I'm afraid that I do think so. – Jan 05 '16 at 06:36
Im thinkin that converting it to $sin(2\theta)$ and $-cos(2\theta)$ will change things – Archis Welankar Jan 05 '16 at 06:40
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Huh? Writing it in a different form will not change the values of the derivatives (unless you mean something else). – Jan 05 '16 at 06:41

score 1 · Accepted Answer · answered Jan 05 '16 at 06:28

No, it is not valid to approximate $\sin^2 x$ by $x^2$, since there are higher order terms that are still significant. Using the expansion

$$\sin x = x - \frac{x^3}{6} + O(x^5)$$

we have that

$$\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$$ so that

$$\frac{x^2 - \sin^2 x}{\tan(3x^4)} = \frac{\frac 1 3 x^4 + O(x^6)}{\tan(3x^4)}$$

leading to an overall limit that is non-zero.

score 1 · Answer 3 · answered Jan 05 '16 at 07:05

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$$L=\lim_{x \to 0}\frac{x^2-sin^2x}{3x^4} \times \lim_{x \to 0} \frac{3x^4}{tan(3x^4)}=\frac{1}{3}\lim_{x \to 0}\frac{x^2-sin^2x}{x^4}$$ $$L=\frac{1}{3}\lim_{x \to 0}\frac{x-sinx}{x^3} \times \lim_{x \to 0}\frac{x+sinx}{x}=\frac{2}{3}\lim_{x \to 0}\frac{x-sinx}{x^3}$$

Now you can use Lhopital's rule

answered Jan 05 '16 at 07:05

Ekaveera Gouribhatla

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It's worth pointing out that this still requires three applications of L'Hospital's rule to evaluate that last limit. – Jan 05 '16 at 07:30
ofcourse my dear user it needs two applications but it hardly takes $10$ seconds,try it once. – Ekaveera Gouribhatla Jan 05 '16 at 07:35

$\lim \frac{x^2-\sin^2{x}}{\tan(3x^4)}$ as $x$ goes to $0$

3 Answers3