# Limit $\lim_{u\to0} \frac{3u}{\tan 2u}$

I’m currently stuck trying to evaluate this limit, $$\lim_{u\to0} \frac{3u}{\tan(2u)},$$ without using L’Hôpital’s rule. I’ve tried both substituting for $$\tan(2u)=\dfrac{2\tan u}{1-(\tan u)^2}$$, and $$\tan 2u=\dfrac{\sin 2u}{\cos 2u}$$ without success. Am I on the right path to think trig sub?

• Are you allowed to use the limit: $\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1.$? – D.B. Nov 23 '18 at 17:31
• Sorry all, limit as u goes to 0. D.B. yes, that trigonometric limit has been covered and can be used. – Isosceles Nov 23 '18 at 17:33
• Why not apply the definition $$\lim_{u \to a}\frac{f(u) - f(a)}{u-a}=f'(a)$$ to $f(u):=\tan(2u)$ and $a:=0$? – Olivier Oloa Nov 23 '18 at 17:34
• Ok. So you can transform your limit into one involving the limit just mentioned by using the fact that $\tan(x) = \sin(x)/\cos(x)$. – D.B. Nov 23 '18 at 17:34
• @YadatiKiran Thanks for your help with editing posts. I will point out that neither \displaystyle nor \dfrac should be used in the titles. For more details, see here: Guidelines for good use of $\rm\LaTeX$ in question titles – Martin Sleziak Nov 23 '18 at 17:36

Note that $$\lim_{u\to 0}\frac{3u}{\tan(2u)}=\frac{3}{2}\times\lim_{u\to 0}\frac{2u}{\sin(2u)}\times\lim_{u\to 0}\cos(2u).$$ Now use your knowledge of well-known limits.

Because of Taylor series, $$\tan2u\sim2u, \quad u\to0$$. Then

$$\lim_{u\to0}\frac{3u}{2u}=\frac32$$

\begin{aligned} \lim_{u\to 0}\frac{3u}{\tan(2u)}&=\lim_{u\to 0}\frac{3u\cos(2u)}{\sin(2u)}\\ &=\lim_{u\to 0}\frac{3\cos(2u)}{2}\cdot\frac{2u}{\sin(2u)}\\ &=\left(\lim_{u\to 0}\frac{3\cos(2u)}{2}\right)\cdot \left(\lim_{u\to 0}\frac{2u}{\sin(2u)}\right)\qquad\text{since both limits exist}\\ &=\frac{3}{2}\cdot 1\\ &=\frac{3}{2}. \end{aligned}

Just Taylor the $$\tan(2u)$$ and the answer comes straight away after you divide by $$u$$ both the numerator and denominator. This is assuming that $$u$$ tends to $$0$$. With infinity limit does not exist.

Hint:

$$\lim_{u\to0} \frac{3u}{\tan(2u)} = \frac{3}{2}\cdot\lim_{u\to0} \bigg[\frac{2u}{\sin(2u)}\cdot\cos (2u)\bigg]$$

Recall that $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

and apply it here.

$$\lim_{x\to 0} {3x\over \tan 2x}=\lim_{x\to 0} {3x\cos 2x\over \sin 2x}=\lim_{x\to 0} {3x\over \sin 2x}={3\over 2}$$

The key point is the strandard limit as $$x\to 0 \,\frac{\sin x}x\to 1$$, indeed we have that

$$\dfrac{3u}{\tan(2u)}=\dfrac{3u}{2u}\dfrac{2u}{\tan(2u)}=\dfrac{3}{2}\dfrac{2u}{\sin(2u)}\cos (2u)\to \frac32\cdot 1 \cdot 1 = \frac32$$

with your first idea we obtain

$$\dfrac{3u}{\tan(2u)}=\dfrac{3u}{2\tan(u)}(1-(\tan u)^2))=\frac32\frac u {\sin u}\cos u(1-(\tan u)^2))\to \frac32\cdot 1\cdot1\cdot 1=\frac32$$

Refer to the related

By MVT

$$\tan(2u)-\tan(0)=$$ $$(2u-0)\Bigl(1+\tan^2(c)\Bigr)=$$

$$2u(1+\tan^2(c))$$

when $$u\to 0, \; c\to 0$$.

the limit is then $$\frac 32$$