Computing limit of $\frac{x-\sin x}{x-\tan x}$ without L'Hôpital

Question

I want to compute following:

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

I have tried to calculate this with l’Hôpital's rule. l’Hôpital's rule states that: $$ \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a } \frac{f'(x)}{g'(x)} $$

Now i can get to the right result with l’Hôpital's rule but it took a little over two pages on paper. Had to use l’Hôpital's rule 4 times. How do you solve this without l’Hôpital's rule ?

The result appears to be(with l’Hôpital's rule):

$$ \lim_{x \rightarrow 0} \frac{x-\sin(x)}{x-\tan(x)}=-\frac{1}{2} $$

If someone can provide alternative solution to this problem that would be highly appreciated.

Answer 1

Utilizing the Taylor expansions

$\sin(x)=x-\frac{1}{6}x^3+\mathcal{O}(x^5)$

$\tan(x)=x+\frac{1}{3}x^3+\mathcal{O}(x^5)$, we get

$\frac{x-\sin(x)}{x-\tan(x)}=\frac{x-(x-\frac{1}{6}x^3+\mathcal{O}(x^5))}{x-(x+\frac{1}{3}x^3+\mathcal{O}(x^5))}=\frac{\frac{1}{6}x^3+\mathcal{O}(x^5)}{-\frac{1}{3}x^3+\mathcal{O}(x^5)}\to-\frac{1}{2}$ as $x\to0$.

Answer 2

By L'Hospital anyway (twice):

$$ \lim_{x \rightarrow 0} \frac{x-\sin(x)}{x-\tan(x)}= \lim_{x \rightarrow 0} \frac{1-\cos(x)}{1-(\tan^2(x)+1)}= \lim_{x \rightarrow 0} \frac{\sin(x)}{-2\tan(x)(\tan^2(x)+1)}.$$

Now you can simplify the sines and evaluate: $-\dfrac12$.

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By L'Hospital, once:

$$\lim_{x \rightarrow 0} \frac{x-\sin(x)}{x-\tan(x)}= \lim_{x \rightarrow 0} \frac{1-\cos(x)}{1-(\tan^2(x)+1)}$$

then using the property $\sin x/x\to1$,

$$\lim_{x \rightarrow 0} \frac{1-\cos(x)}{-\tan^2(x)}=\lim_{x \rightarrow 0} \frac{2\sin^2\left(\dfrac x2\right)}{-\dfrac{\sin^2(x)}{\cos^2(x)}}=\lim_{x \rightarrow 0} \frac{2\left(\dfrac x2\right)^2}{-x^2}.$$

Answer 3

Using l'Hoptial's rule repeatedly is sometimes necessary. But frequently, after only one or two applications, one can use algebra and trigonometric identities to simplify and then evaluate the limit. Starting with $$ L = \lim_{x \rightarrow 0} \frac{x - \sin x}{x - \tan x} $$ we verify that we have the indeterminate form $\left[ \frac{0}{0} \right]$, so application of l'Hopital's rule is valid. We get \begin{align*} L &\overset{l'H}= \lim_{x \rightarrow 0} \frac{1 - \cos x}{1 - \sec^2 x} \\ &= \lim_{x \rightarrow 0} \frac{(1 - \cos x)(\cos^2 x)}{(1 - \sec^2 x)(\cos^2 x)} \\ &= \lim_{x \rightarrow 0} \frac{(1 - \cos x)(\cos^2 x)}{\cos^2 x - 1} \\ &= \lim_{x \rightarrow 0} \frac{(1 - \cos x)(\cos^2 x)}{-\sin^2 x} \\ &= \lim_{x \rightarrow 0} \frac{(1 - \cos x)(\cos^2 x)(1 + \cos x)}{(-\sin^2 x)(1 + \cos x)} \\ &= \lim_{x \rightarrow 0} \frac{(1 - \cos^2 x)(\cos^2 x)}{(-\sin^2 x)(1 + \cos x)} \\ &= \lim_{x \rightarrow 0} \frac{(\sin^2 x)(\cos^2 x)}{(-\sin^2 x)(1 + \cos x)} \\ &= \lim_{x \rightarrow 0} \frac{-\cos^2 x}{1 + \cos x} \\ &= \frac{-1}{2} \text{.} \end{align*} All the subsequent manipulations are standard trigonometric manipulations and identities. (... shown in excruciating detail. Normally, my students would show about half as many steps.) The initial form is indeterminant because the numerator and denominator both "have a root" at $x=0$. The subsequent manipulations isolate that root into a factor in the denominator and then arrange to cancel it with a factor in the numerator. The result no longer is a $\left[ \frac{0}{0} \right]$ form, so we can just evaluate the nice, continuous result at $0$ to get the value of the limit.

So we can apparently avoid a couple of pages of fussy derivatives by recognizing that we can finish by algebra and identities after only one step. What if we hadn't noticed that after one step and had applied l'Hopital a second time...\begin{align*} L &\overset{l'H}= \lim_{x \rightarrow 0} \frac{\sin x}{-2 \sec^2 x \tan x} \\ &= \lim_{x \rightarrow 0} \frac{(\sin x)(\cos^3 x)}{(-2 \sec^2 x \tan x)( \cos^3 x)} \\ &= \lim_{x \rightarrow 0} \frac{\sin x \cos^3 x}{-2 \sin x} \\ &= \lim_{x \rightarrow 0} \frac{\cos^3 x}{-2 } \\ &= \frac{-1}{2} \end{align*} Again, we use a little trigonometry and are done quickly.