MHB 2.1.207 AP calculus practice exam problem Lim with tan(4X)/6x

[karush]

$\displaystyle\lim_{{x}\to{0}}\left(\frac{\tan 4x}{6x}\right)=$

(A) $\dfrac{1}{3}$

(B) $\dfrac{2}{3}$

(C) 0

(D) $-\dfrac{2}{3}$

(E) DNE

solution
[sp]
direct substitution of 0 results in undeterminant so use LH'R
so then after taking d/dx of numerator and denominator and factor out constant we have

$\displaystyle\lim_{{x}\to{0}}\left(\frac{4\sec ^2\left(4x\right)}{6}\right)
=\dfrac{4}{6} \lim_{{x}\to{0}} \sec ^2(4x) $
take the limit then simplify then

$\dfrac{4}{6}\cdot 1=\dfrac{2}{3}\quad (B)$[/sp]

ok hopefully correct probable need some more itermeddiate steps
I added show/hide

 

[Opalg]

$\displaystyle\lim_{{x}\to{0}}\left(\frac{\tan 4x}{6x}\right)=$

(A) $\dfrac{1}{3}$

(B) $\dfrac{2}{3}$

(C) 0

(D) $-\dfrac{2}{3}$

(E) DNE

solution
[sp]
direct substitution of 0 results in undeterminant so use LH'R
so then after taking d/dx of numerator and denominator and factor out constant we have

$\displaystyle\lim_{{x}\to{0}}\left(\frac{4\sec ^2\left(4x\right)}{6}\right)
=\dfrac{4}{6} \lim_{{x}\to{0}} \sec ^2(4x) $
take the limit then simplify then

$\dfrac{4}{6}\cdot 1=\dfrac{2}{3}\quad (B)$[/sp]

ok hopefully correct probable need some more itermeddiate steps
I added show/hide

Looks good to me. (Rock)

 

[Greg]

$$\frac{4x}{4x}\cdot\frac{\tan4x}{6x}=\frac{4x}{4x}\cdot\frac{\sin4x}{6x\cdot\cos{4x}}=\frac{4x}{6x}\cdot\frac{\sin4x}{\cos4x\cdot4x}$$

$$\frac23\lim_{x\to0}\frac{\sin4x}{\cos4x\cdot4x}=\frac23$$