Question

X---> lim pi whole root(5+cosx) -2 / (pi-x)^2 tough one *_*

Answer 1

Answer:

$\frac{1}{8}$

Step-by-step explanation:

$\lim_{x \to \\pi } \frac{\sqrt{5+cosx}-2}{(\pi -x)^{2}} * [\frac{\sqrt{5+cosx} + 2 }{\sqrt{5+cosx} + 2} ]$

$\lim_{x \to \\pi }  \frac{5 + cosx - 4}{(\pi-x)^{2} } * \frac{1}{\sqrt{5+cosx}+2 }$

$\lim_{x \to \\pi } \frac{1+cosx}{(\pi -x)^{2} } * \frac{1}{\sqrt{5+cosx} +2}$

Now,

Let x - $\pi$ = S

or, x = $\pi + S$

So,

$\lim_{S \to 0} \frac{1+cos(\pi +S)}{(\pi-\pi-S )^{2} } * \frac{1}{\sqrt{5+cos(\pi +S)} +2 }$

$\lim_{S \to 0} \frac{1-cos S}{S^{2} } * \frac{1}{\sqrt{5-cosS}+2 }$

$\lim_{S \to 0} 2sin^{2} \frac{S}{2} * \frac{1}{S^{2}} * \frac{1}{\sqrt{5-cosS}+2 }$

$\lim_{S \to 0} 2\frac{sin^{2}\frac{S}{2}  }{\frac{S^{2} }{4} } * \frac{1}{S^{2} } * \frac{S^{2} }{4} * \frac{1}{\sqrt{5-cosS} +2 }$

We can see from here,

(sin x / x)^2 = 1 , where x = S/2

$\lim_{S \to 0} \frac{1}{\sqrt{5-cosS}+2 } * \frac{1}{2}$

Now, putting the limit,

$\frac{1}{\sqrt{5-1} +2} * \frac{1}{2}$     [as cos0 = 1]

$\frac{1}{2+2} * \frac{1}{2}$

$\frac{1}{8}$

Thanks !

#BAL #answerwithquality