Ytukyg

I’m currently stuck trying to evaluate this limit,
$$
lim_{uto0} frac{3u}{tan(2u)},
$$
without using L’Hôpital’s rule. I’ve tried both substituting for $tan(2u)=dfrac{2tan u}{1-(tan u)^2}$, and $tan 2u=dfrac{sin 2u}{cos 2u}$ without success. Am I on the right path to think trig sub?

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

Because of Taylor series, $tan2usim2u, quad uto0$. Then

$$lim_{uto0}frac{3u}{2u}=frac32$$

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

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:

$$lim_{uto0} frac{3u}{tan(2u)} = frac{3}{2}cdotlim_{uto0} bigg[frac{2u}{sin(2u)}cdotcos (2u)bigg]$$

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

and apply it here.

By MVT

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

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

when $uto 0, ; cto 0$.

the limit is then
$$frac 32$$

$$lim_{xto 0} {3xover tan 2x}=lim_{xto 0} {3xcos 2xover sin 2x}=lim_{xto 0} {3xover sin 2x}={3over 2}$$

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

$$dfrac{3u}{tan(2u)}=dfrac{3u}{2u}dfrac{2u}{tan(2u)}=dfrac{3}{2}dfrac{2u}{sin(2u)}cos (2u)to frac32cdot 1 cdot 1 = frac32$$

with your first idea we obtain

$$dfrac{3u}{tan(2u)}=dfrac{3u}{2tan(u)}(1-(tan u)^2))=frac32frac u {sin u}cos u(1-(tan u)^2))to frac32cdot 1cdot1cdot 1=frac32$$

Refer to the related

• How to prove that $limlimits_{xto0}frac{sin x}x=1$?