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Find $$\lim_{x \to 0} \frac{\tan x - x}{x^3} $$

So I know that for L'Hôpital's rule I take the derivative of the top and the bottom, and I keep doing that until I get something that isn't an indeterminate form. So I got $\dfrac{\sec^2(x)-1}{3x^2}$ and I still couldn't plug in $0$ so I took the derivatives again and got $\dfrac{2\tan(x)\sec^2(x)}{6x}$ and still can't plug in $0$. Do I have to take the derivatives again and use product rule for the numerator? Thanks for the help!

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    $\begingroup$ @Chokladkakan, you changed the limit to $\infty$ in place of $0$ $\endgroup$ – lab bhattacharjee Mar 6 '13 at 18:27
  • $\begingroup$ @labbhattacharjee Oh my! Thank you for noticing! $\endgroup$ – tesc Mar 6 '13 at 18:30
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$$\frac{\sec^2x-1}{3x^2}=\frac{\tan^2x}{3x^2}=\frac13\left(\frac{\tan x}x\right)^2$$

Now, $\lim_{x\to0}\frac{\tan x}x=\lim_{x\to0}\sec^2x=1$

Also, $\{\lim_{x\to0}f(x)\}^n=\lim_{x\to0}\{f(x)\}^n$ for positive integer $n$

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  • $\begingroup$ oh I see! completely forgot that identity, thanks! $\endgroup$ – user56852 Mar 6 '13 at 18:22
  • $\begingroup$ @user56852, trigonometric identities I felt, indispensable for limit and integration. $\endgroup$ – lab bhattacharjee Mar 6 '13 at 18:30
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I wouldn't take l'hospitale in your case, $$\tan(x)=x+\frac{1}{3} x^3+ \frac{2}{15}x^5 + \dots, $$ So $$\lim_{x\to 0} \frac{\tan(x)-x}{x^3}= \lim_{x\to 0} \frac{\frac{1}{3} x^3+ \frac{2}{15} x^5+ \dots}{x^3}= \lim_{x\to 0} \frac{1}{3} + \frac{2}{15}x^2 + \dots $$

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