# Limit: $\lim_{x\to 0} \frac {x-\sin x}{x-\tan x}$

Here's the question: $$\lim_{x\to 0} \frac {x-\sin x}{x-\tan x}$$

I've used l'Hospitals to get $$\lim_{x\to 0} \frac {1-\cos x}{1-\sec^2x}$$ I then tried to use it again, resulting in $= \lim\limits_{x\to 0} \cfrac {\sin x}{1-\sec^2 x\cdot\tan^2 x}$ which gives me 0, but the wolfram alpha answer is $-1/2$. I've tried other things like dividing by x, but nothing I get leads me to believe I'm on the right path.

• $\dfrac {\operatorname{d}}{\operatorname{d}x}(1-\sec^2 x) = -2 \tan x \sec^2 x$ – Graham Kemp Mar 26 '14 at 16:28
• See this answer math.stackexchange.com/a/438121/72031 Equation (7) of that answer is this limit in question. robjohn has done it without the use of L'Hospital. – Paramanand Singh Mar 27 '14 at 3:38

Yes, firt use the l'Hospital rule: $$\lim_{x\rightarrow 0}\frac{x-\sin x}{x-\tan x}=\lim_{x\rightarrow 0}\frac{1-\cos x}{1-\sec^2x}$$ Now, $\sec^2x=\frac{1}{\cos^2x}$, then $$\lim_{x\rightarrow 0}\frac{x-\sin x}{x-\tan x}=\lim_{x\rightarrow 0}\frac{1-\cos x}{1-\frac{1}{\cos^2x}}=\lim_{x\rightarrow 0}\frac{1-\cos x}{\frac{\cos^2x-1}{\cos^2x}} =\lim_{x\rightarrow 0}(1-\cos x)\frac{\cos^2x}{\cos^2x-1}=$$ $$=\lim_{x\rightarrow 0} (1-\cos x)\frac{\cos^2x}{-(1-\cos x)(1+\cos x)}=$$ $$=\lim_{x\rightarrow 0}-\frac{\cos^2x}{1+\cos x}=-\frac{1}{2}.$$

Hint $\dfrac{1-\cos x}{1-\sec^2x} = \dfrac{1-\cos x}{1-\sec^2x} \cdot \dfrac{\cos^2 x}{\cos^2 x}= \dfrac{(1-\cos x)\cos^2 x}{\cos^2 x-1} = \dfrac{(1-\cos x)\cos^2 x}{-1(1-\cos^2 x)}$

• Doesn't cos x = 1 at x=0? So then the last step would be (1-1)*1/-1*(1-1) = 0 would it not? – Greener Mar 27 '14 at 12:10
• @Greener I didn't finish the problem. If you factor and cancel in the last step, you remove that issue and get the correct answer. – John Habert Mar 27 '14 at 13:48
• Okay, I didn't see that until later. Thanks for the help. – Greener Mar 27 '14 at 16:07

After applying l'Hospital twice you get: $$\lim_{x\to 0} \frac {\sin x}{\frac{-2\sin x\cos x}{\cos^4 x}}$$, which tends to -1/2

HINT By the Maclaurin expansion of $\sin x$ and $\tan x$, one have

$$\lim_{x \to 0}\frac{x-\sin x}{x-\tan x}=\lim_{x \to 0}\frac{\frac{1}{6}x^3-\frac{1}{120}x^5+\mathcal{o}(x^5)}{-\frac{1}{3}x^3-\frac{2}{15}x^5+\mathcal{o}(x^5)}$$