- #1

karush

Gold Member

MHB

- 3,269

- 5

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

(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

(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

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