Question

$let \: f(x)$
$= \frac{1 - \cos(4x) }{ {x}^{2} } \: \: \: \: \: \: \: \: \: \: \: x < 0$

$= a \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: x = 0$
$= \frac{ \sqrt{x} }{ \sqrt{16 + \sqrt{x} } - 4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x > 0$


Determine the value of 'a' if possible, so that the function is continuous at x = 0.



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Answer 1

Answer:

a = 8

Step-by-step explanation:

(i) When x < 0: LHL

$\lim_{x \to 0} \frac{1 - cos^4x}{x^2}$

$=>\lim_{x \to 0} \frac{2sin^22x}{x^2}$

$=>2{[\lim_{x \to 0} \frac{sin2x}{2}}]^2*\frac{4x^2}{x^2}$

$=>2 * 1 * 4$

$=>8$

(ii) When x = 0:

Given, f(x) is continuous at x = 0.

f(0) = a = LHL

∴ a = 8

(iii) When x > 0: RHL

$=\frac{\sqrt{x}}{\sqrt{16} + \sqrt{x}-4}$

$=\lim_{x \to 0} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} * \frac{\sqrt{16+\sqrt{x}}+4}{\sqrt{16+\sqrt{x}}+4}$

$=\lim_{x \to 0} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16}$

$=\lim_{x \to 0} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}}$

$=\lim_{x \to 0} \sqrt{16+\sqrt{x}+4}$

$=\sqrt{16} + 4$

$=8$

From (i) & (ii) & (iii), we get

a = 8.

Hope it helps!

Answer 2

ANSWER:--------------

=>$2[lim⁡x →[tex]0sin2x2]2∗4x2x2 =>2{[\lim_{x \to 0} \frac{sin2x}{2}}]^2*\frac{4x^2}{x^2}=>2[limx$

→$02sin2x]2∗x24x2 =>2∗1∗4=>2 * 1*4 =>2∗1∗4 =>8=>8=>8$

x = 0:

at x = 0.

f(0) = a = LHL

note here now

∴ a = 8

(x > 0: RHL

$=x16+x−4=\frac{\sqrt{x}}{\sqrt{16}+\sqrt{x}-4}=16+x−4x=lim⁡x$

→$0x16+x−4∗16+x+416+x+4=\lim_{x \to 0} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$

$\frac{\sqrt{16+\sqrt{x}}+4}{\sqrt{16+\sqrt{x}}+4}=limx$

→$016+x−4x∗16+x+416+x+4=lim⁡x$

→$0x(16+x+4)16+x−16=\lim_{x \to 0} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16}=limx$

→$016+x−16x(16+x+4)=lim⁡x$

→$0x(16+x+4)x=\lim{x \to0}\frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}$

${\sqrt{x}}=limx →0xx(16+x+4)=lim⁡x →016+x+4=\lim_{x \to 0}$

[tex\sqrt{16+\sqrt{x}+4}=limx[/tex]

→$016+x+4=16+4=\sqrt{16} +4=16+4=8=8=8$

from all thus steps ,thus we get a=8

a = 8.

hence proved..

hope it helps:--

T!—!ANKS!!!!