Evaluating $\lim_{x\to \pi/2}\frac{\tan11x-\tan7x}{\tan9x+\tan13x}$ without L'Hopital

Question

I want to know that how a student can find the limit of $$\lim_{x\to \pi/2}\frac{\tan(11x)-\tan(7x)}{\tan(9x)+\tan(13x)}$$ without using L-Hospital's rule.

My Way:

What I observed is that the following limit is of the form of $\frac{\infty}{\infty}$. Now I know how to solve this types of limits.

For e.g. : $$\lim_{x\to \infty}\frac{x^{2}+x+4}{x^{2}+3x+7}=1.$$

Now there are two ways of doing this. One is by dividing the numerator and denominator by $x^{2}$ and the other one is by applying L-Hospital's rule.

But in my integral I think one can apply only L-Hospital's rule. But I need any other rule. Is that possible?

If I also apply the formula $\tan(A)-\tan(B)=\tan(A-B)(1+\tan(A)\tan(B))$

then also I am not getting any tricks.

Please help me out with this limit.

Answer 1

This is the same as

$$\lim_{x\to0}\frac{\tan(11x+\frac\pi2)-\tan(7x+\frac\pi2)}{\tan(9x+\frac\pi2)+\tan(13x+\frac\pi2)}=\lim_{x\to0}\frac{\cot(11x)-\cot(7x)}{\cot(9x)+\cot(13x)}.$$

Now using $\lim_{t\to0}t\cot(t)=1$, we can replace $\cot(t)$ by $\dfrac1t$, and simplify this gives

$$\frac{\frac1{11}-\frac17}{\frac19+\frac1{13}}.$$

---

Addendum:

As kindly given by Wolfram Alpha,

$$\cot(7x)=\frac{-1 + 2 \cos(2 x) - 2 \cos(4 x) + 2 \cos(6 x)}{1 + 2 \cos(2 x) + 2 \cos(4 x) + 2 \cos(6 x)}\cot(x)\sim \frac{\cot(x)}7$$ and similar identities for other coefficients. Besides the factor $\cot(x)$, the numerator of the fraction always tends to $1$ and the denominator to the coefficient. So the replacement with $\cot(t)\leftrightarrow\dfrac1t$ is unnecessary.

Answer 2

We have, \begin{align*} \lim_{x \to \frac{\pi}{2}} \frac{\tan(11x) -\tan(7x)}{\tan(9x) + \tan(13x)} & = \lim_{x \to \frac{\pi}{2}} \frac{\sin(11x)\cos(7x) - \sin(7x)\cos(11x)}{\sin(9x)\cos(13x) + \sin(13x)\cos(9x)} \cdot \frac{\cos(9x)\cos(13x)} {\cos(7x)\cos(11x)}\\ & = \lim_{x \to \frac{\pi}{2}} \frac{\sin(4x)}{\sin(22x)} \cdot \frac{\cos(9x)\cos(13x)} {\cos(7x)\cos(11x)}\\ & = \lim_{h \to 0} \frac{\sin(2\pi + 4h)}{\sin(11\pi + 22h)} \cdot \frac{\cos(9\pi/2 + 9h)\cos(13\pi/2 + 13h)} {\cos(7\pi/2 + 7h)\cos(11\pi/2 + 11h)} \\ & = \lim_{h \to 0} \frac{\sin(4h)}{-\sin(22h)} \cdot \frac{\sin( 9h)\sin(13h)} {\sin(7h)\sin(11h)} \\ & = -\frac{4}{22} \cdot \frac{9} {7} \cdot \frac{13} {11}. \end{align*}

In the third step, we used the substitution $x = \pi/2 + h$.

Answer 3

After Yves's beautiful solution: $$\lim_{x\to \pi/2}\frac{\tan(11x)-\tan(7x)}{\tan(9x)+\tan(13x)}\\=\lim_{x\to \pi/2}\frac{(x-\tfrac\pi 2)\tan(11x)-(x-\tfrac\pi 2)\tan(7x)}{(x-\tfrac\pi2)\tan(9x)+(x-\tfrac\pi2)\tan(13x)}\\ =\frac{-\tfrac1{11}-\tfrac17}{-\tfrac19-\tfrac1{11}}=-\frac{4.13.9}{11.7.22}$$ due to the amazing limit $$\lim_{x\to\frac\pi2}(x-\tfrac\pi2)\tan(2k+1)x=-\frac1{2k+1}.$$