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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?

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  • $\begingroup$ Are you allowed to use the limit: $\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1.$? $\endgroup$ – D.B. Nov 23 '18 at 17:31
  • $\begingroup$ Sorry all, limit as u goes to 0. D.B. yes, that trigonometric limit has been covered and can be used. $\endgroup$ – Isosceles Nov 23 '18 at 17:33
  • $\begingroup$ Why not apply the definition $$ \lim_{u \to a}\frac{f(u) - f(a)}{u-a}=f'(a) $$ to $f(u):=\tan(2u)$ and $a:=0$? $\endgroup$ – Olivier Oloa Nov 23 '18 at 17:34
  • $\begingroup$ Ok. So you can transform your limit into one involving the limit just mentioned by using the fact that $\tan(x) = \sin(x)/\cos(x)$. $\endgroup$ – D.B. Nov 23 '18 at 17:34
  • $\begingroup$ @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 $\endgroup$ – Martin Sleziak Nov 23 '18 at 17:36
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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.

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

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

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

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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.

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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.

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

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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

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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 $$\frac 32$$

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