$9$

#### Explanation:

$\setminus {\lim}_{x \setminus \to 0} \setminus \frac{\setminus \ln \left(1 + 9 x\right) + \setminus \cos \left(4 x\right) - 1}{x}$

Applying L'Hospital's rule for $\frac{0}{0}$ form as follows

$\setminus {\lim}_{x \setminus \to 0} \setminus \frac{\setminus \frac{d}{\mathrm{dx}} \left(\setminus \ln \left(1 + 9 x\right) + \setminus \cos \left(4 x\right) - 1\right)}{\setminus \frac{d}{\mathrm{dx}} x}$

$= \setminus {\lim}_{x \setminus \to 0} \setminus \frac{\setminus \frac{1}{1 + 9 x} \setminus \cdot 9 - 4 \setminus \sin \left(4 x\right)}{1}$

$= \setminus {\lim}_{x \setminus \to 0} \left(\setminus \frac{9}{1 + 9 x} - 4 \setminus \sin \left(4 x\right)\right)$

$= \setminus \frac{9}{1 + 0} - 4 \left(0\right)$

$= 9$

Jun 28, 2018

$9$.

#### Explanation:

We will use the following Standard Limits :

$\left(1\right) : {\lim}_{t \to 0} {\left(1 + t\right)}^{\frac{1}{t}} = e$.

$\left(2\right) : {\lim}_{h \to 0} \frac{\sinh}{h} = 1$.

$\text{Now, the Reqd. Lim.} = {\lim}_{x \to 0} \frac{\ln \left(1 + 9 x\right) + \cos 4 x - 1}{x}$,

$= \lim \frac{1}{x} \ln \left(1 + 9 x\right) - \frac{1 - \cos 4 x}{x}$,

$= \lim \left(9 \cdot \frac{1}{9 x}\right) \ln \left(1 + 9 x\right) - \frac{2 {\sin}^{2} 2 x}{x}$,

$= \lim \left[9 \cdot \ln {\left(1 + 9 x\right)}^{\frac{1}{9 x}} - 2 \cdot {\left(2 \sin x \cos x\right)}^{2} / x\right]$,

$= \lim \left[9 \cdot \ln {\left(1 + 9 x\right)}^{\frac{1}{9 x}} - 8 \cdot \sin \frac{x}{x} \cdot \sin x {\cos}^{2} x\right]$.

Since, $\ln$ function is continuous, we have,

${\lim}_{x \to 0} \ln {\left(1 + 9 x\right)}^{\frac{1}{9 x}} = \ln \left\{{\lim}_{x \to 0} {\left(1 + 9 x\right)}^{\frac{1}{9 x}}\right\} = \ln e$,

$= 1$.

Hence, $\text{the reqd. lim} = 9 \cdot 1 - 8 \cdot 1 \cdot \sin 0 \cdot {\cos}^{2} 0$,

$= 9 - 0$,

$= 9$.

Enjoy Maths.!