If limx→0 2ax+(a−1)sin⁡xtan3⁡x=l, then a+l is equal to

A
$\frac{2}{3}$
B
$\frac{1}{3}$
C
$\frac{2}{9}$
D
$\frac{4}{9}$

Solution:

We have

$\begin{array}{l} lim_{x \to 0} \frac{2 a x + (a - 1) \sin x}{\tan^{3} x} = l \\ \Rightarrow lim_{x \to 0} \frac{2 a + (a - 1) \frac{^{\sin x}}{x}}{x^{2} {(\frac{\tan x}{x})}^{3}} = l \end{array}$

We observe that the numerator on LHS tends to $3 a - 1$ and the denominator tends to 0. So, the limit will be a finite number $I$
only if

$3 a - 1 = 0$ i.e., $a = \frac{1}{3}$

$∴ lim_{x \to 0} \frac{3 (1 - \sin x)}{x^{2} {(\frac{\tan x}{x})}^{3}} = 1$

$\Rightarrow lim_{x \to 0} \frac{\frac{x - \sin x}{x^{3}}}{{(\begin{matrix} \tan x \\ x \end{matrix})}^{3}} = \frac{3 l}{2}$

$\Rightarrow \frac{1}{6} = \frac{3 l}{2} \Rightarrow l = \frac{1}{9} [∵ lim_{x \to 0} \frac{x - \sin x}{x^{3}} = \frac{1}{6}]$

Hence, $a + l = \frac{1}{3} + \frac{1}{9} = \frac{4}{9}$

If $lim_{x \to 0} \frac{2 a x + (a - 1) s i n x}{t a n^{3} x} = l,$ then $a + l$ is equal to

Solution:

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