Question

plz solve part 6 of this question​

Answer 1

Answer:

$\large\bold\red{4 \sqrt{2} \times ln(5) \times ln(3) }$

Step-by-step explanation:

Given,

$(vi) \lim_{x\to0}</p><p> \: \frac{ {15}^{x} - {5}^{x} - {3}^{x} + 1}{ \sqrt{2} - \sqrt{1 + \cos(x) } }$

After simplifying,

We get,

$= \lim_{x\to0} \frac{ {5}^{x} ( {3}^{x} - 1) - 1( {3}^{x} - 1) }{ \sqrt{2} - \sqrt{1 + \cos(x) } }$

But,

we know that,

$\bold{1 + \cos(2 \alpha ) = 2 { \cos}^{2} \alpha }$

Therefore,

we get,

$= \lim_{x\to0}</p><p> \frac{( {5}^{x} - 1)( {3}^{x} - 1)}{ \sqrt{2} (1 - \cos\frac{x}{2} ) } </p><p>$

Now,

multiplying and dividing the numerator and denominator by ${x}^{2}$,

we get,

$= \lim_{x\to0} (\frac{ {5}^{x} - 1}{x}) ( \frac{ {3}^{x} - 1 }{x}) ( \frac{ {x}^{2} }{ \sqrt{2} (1 - \cos \frac{x}{2} ) } )</p><p>$

But,

we know that,

$\bold{\lim_{x\to0} (\frac{ {a}^{x} - 1 }{x} ) = ln(a)} \\ \\ and \\ \\\bold{ 1 - \cos( \frac{x}{2} ) = 2 { \sin}^{2} \frac{x}{4} }$

Therefore,

Applying the formula and pUtting the values,

we get,

$= ln(5) ln(3) \lim_{x\to0}</p><p> \frac{ {x}^{2} }{2 \sqrt{2} { \sin}^{2} \frac{x}{4} } \\ \\ = ln(5) ln(3) \lim_{x\to0}</p><p> \frac{ {x}^{2} }{2 \sqrt{2} {( \sin \frac{x}{4} ) }^{2} } \\ \\ = ln(5) ln(3) \lim_{x\to0}</p><p> \frac{ {x}^{2} }{2 \sqrt{2} \frac{ {( \sin \frac{x}{4} ) }^{2} }{ {( \frac{x}{4}) }^{2} } \times {( \frac{x}{4}) }^{2} }$

But,

we know that,

$\bold{\lim_{x\to0} \frac{ \sin(x) }{x} = 1}</p><p>$

Therefore,

we get,

$= ln(5) ln(3) \lim_{x\to0}</p><p> \frac{ {x}^{2} }{2 \sqrt{2} \times \frac{ {x}^{2} }{16} } \\ \\ = \frac{16 \times ln(5) \times ln(3) }{2 \sqrt{2} } \\ \\ = \bold{4 \sqrt{2} \times ln(5) \times ln(3) }$