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I am trying to use L'Hopital's rule, but I'm not sure if I'm doing this correctly. In my textbook, the rule is stated as follows:

$(f(a) = g(a) = 0) \Rightarrow$ $\lim_{x \to a} \frac{f(a)}{g(a)} = \frac{f'(a)}{g'(a)}$ (given $\frac{f'(a)}{g'(a)}$ is defined)

So in this case, I have $f(x) = x - \tan(x)$ and $g(x) = x- \sin(x)$, both of which are $0$ at $x = 0$. Also, $f'(x) = 1 - \sec^2(x)$ and $g'(x) = 1 - \cos(x)$. However, wouldn't this mean the RHS isn't defined, since $\frac{f'(a)}{g'(a)} = \frac{0}{0}$? I am confused how to resolve this.

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    $\begingroup$ So apply L'Hospital's rule again (and again once more!). Both numerator and denominator vanish to third order. $\endgroup$ – user296602 Apr 24 '18 at 18:32
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$\lim_{x \to 0} \frac{x - tan(x)}{x - sin(x)}$

This is a $\frac00$ indeterminate form , hence L'hospitals rule is applicable;

$=\lim_{x\to0}\frac{1-\sec^2(x)}{1-\cos(x)}$

applying L'Hospitals rule once again gives;

$=\lim_{x\to0}\frac{-2\sec^2(x)\tan(x)}{\sin(x)}$

$=\lim_{x\to0}\frac{-2\sec^3(x)\cdot\sin(x)}{\sin(x)}$

$=\lim_{x\to0}-2\sec^3(x)$

$=-2$

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Since

$${1-\sec^2x\over1-\cos x}={1-{1\over\cos^2x}\over1-\cos x}={\cos^2x-1\over\cos^2x(1-\cos x)}={(\cos x+1)(\cos x-1)\over\cos^2x(1-\cos x)}=-{\cos x+1\over\cos^2x}$$

It suffices to L'Hopitate just once:

$$\lim_{x\to0}{x-\tan x\over x-\sin x}=\lim_{x\to0}{1-\sec^2x\over1-\cos x}=-\lim_{x\to0}{\cos x+1\over\cos^2x}=-2$$

In other words, it doesn't hurt to try a little housecleaning after you've taken a bunch of derivatives.

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Differentiating numerator and denominator with respect to $x$ we get $$-\frac{1-(1+\tan(x)^2)}{1-\cos(x)}$$ and once more $$\frac{-2\tan(x)(1+\tan^2(x))}{\sin(x)}$$ and this tends to $-2$ for $x$ tends to zero

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