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I'm calculating the limit for the following equation, and would like some feedback on my solution. The answer is $-2$ although, I cannot seem to derive this whilst using l'hopitals rule.

$\lim_{x \to 0}\frac{x-\tan(x)}{x-\sin(x)}$

my working out:

$$\lim_{x \to 0}\frac{\frac{d}{dx}(x-\tan(x))}{\frac{d}{dx}(x-\sin(x))}=\frac{1-\sec^2(x)}{1-\cos(x)} = \frac{\tan^2(x)(1+\cos(x))}{1-\cos^2(x)}=0$$

Meilton
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  • What makes you think that the limit is zero? The last expression is still of the form $0/0$ for $x \to 0$. – Martin R Mar 01 '21 at 12:53
  • How did you get $=0$, and where did the minus sign from $1-\sec^2 x=-\tan^2 x$ go? – user10354138 Mar 01 '21 at 12:54
  • It seems OK (except for wrting $=$ for things that are not equal) up to $\frac{\tan^2(x)(1+\cos(x))}{1-\cos^2(x)}$, but that still has indeterminate form $0/0$. – GEdgar Mar 01 '21 at 12:54
  • Hint: you divided by zero. Just keep applying lhopital – Steve S Mar 01 '21 at 12:54
  • Also: https://math.stackexchange.com/q/1516339/42969, https://math.stackexchange.com/q/508733/42969. – Martin R Mar 01 '21 at 12:56
  • Aha! I understand it properly now. I was afraid of trying to apply l'hospitals constantly, although the comments make it far clearer. Thank you! – Meilton Mar 01 '21 at 12:59

1 Answers1

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Hint:

The Maclaurin series for tan(x) is:

\begin{equation*} \tan x=x+\frac{x^{3}}{3} +\frac{2x^{5}}{15} +... \end{equation*} And that for sin(x) is \begin{gather*} \sin x=x-\frac{x^{3}}{3!} +\frac{x^{5}}{5!} +...\\ \end{gather*} Can you use this data to get to the answer?

Ishraaq Parvez
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