undersetx→ 0 lim sin^2(π cos^4x) x^4 is equal to :

Given that

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}(\pi {(1-{\mathrm{sin}}^{2}x)}^{2})}{{x}^{4}}$$x→0lim​x4sin2(π(1−sin2x)2)​

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}(\pi (1+{\mathrm{sin}}^{4}x-2{\mathrm{sin}}^{2}x))}{{x}^{4}}$$=x→0lim​x4sin2(π(1+sin4x−2sin2x))​

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}(\pi -\pi (2{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{4}x))}{{x}^{4}}$$=x→0lim​x4sin2(π−π(2sin2x−sin4x))​

$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}(\pi (2{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{4}x))}{{x}^{4}}$=x→0lim​x4sin2(π(2sin2x−sin4x))​$=\underset{x\rightarrow 0}{\mathrm{lim}}{(\frac{\mathrm{sin}[\pi (2{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{4}x)]}{\pi (2{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{4}x)})}^{2}\cdot \frac{{\pi }^{2}{(2{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{4}x)}^{2}}{{x}^{4}}$=x→0lim​(π(2sin2x−sin4x)sin[π(2sin2x−sin4x)]​)2⋅x4π2(2sin2x−sin4x)2​

When $x\rightarrow 0\Rightarrow 2{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{4}x\rightarrow 0$x→0⇒2sin2x−sin4x→0

Let $k=2{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{4}x$k=2sin2x−sin4x

$$=\underset{k\rightarrow 0}{\mathrm{lim}}{[\frac{\mathrm{sin}\pi k}{\pi k}]}^{2}.\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{\pi }^{2}.{\mathrm{sin}}^{4}x{(2-{\mathrm{sin}}^{2}x)}^{2}}{{x}^{4}}$$=k→0lim​[πksinπk​]2.x→0lim​x4π2.sin4x(2−sin2x)2​

We know that $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}kx}{x}=k.$x→0lim​xsinkx​=k.

$$=\frac{{\pi }^{2}}{{\pi }^{2}}.{\pi }^{2}.\underset{x\rightarrow 0}{\mathrm{lim}}{[\frac{\mathrm{sin}x}{x}]}^{4}.\underset{x\rightarrow 0}{\mathrm{lim}}{(2-{\mathrm{sin}}^{2}x)}^{2}$$=π2π2​.π2.x→0lim​[xsinx​]4.x→0lim​(2−sin2x)2

$$={\pi }^{2}.1.{(2)}^{2}$$=π2.1.(2)2

$$=4{\pi }^{2}.$$=4π2.