\(\begin{array}{l}
5)\quad L = \lim\limits_{x\to 0}\dfrac{x}{\sqrt{x+4} - 2}\\
\Leftrightarrow L = \lim\limits_{x\to 0}\dfrac{x\left(\sqrt{x+4} + 2\right)}{\left(\sqrt{x+4} - 2\right)\left(\sqrt{x+4} + 2\right)}\\
\Leftrightarrow L = \lim\limits_{x\to 0}\dfrac{x\left(\sqrt{x+4} + 2\right)}{x}
\Leftrightarrow L = \lim\limits_{x\to 0}\left(\sqrt{x+4} + 2\right)\\
\Leftrightarrow L = \sqrt{0+4} + 2\\
\Leftrightarrow L = 4\\
6)\\
\qquad f(x)= \dfrac{1}{1+\sin^2x}\\
+)\quad f(0)= 1\\
+)\quad f'(x)= - \dfrac{sin2x}{(1+\sin^2x)^2}\\\Rightarrow f'(0)= 0\\
+)\quad f''(x)= \dfrac{4(6\cos2x +\cos4x -3 )}{(cos2x -3)^3}\\\Rightarrow f''(0)= -2\\
+)\quad f'''(x)= \dfrac{8\sin2x(24\cos2x +\cos4x +7)}{(\cos2x -3)^4}\\ \Rightarrow f'''(0)= 0\\
+)\quad f^{(4)}(x)= -\dfrac{4(-66\cos2x + 320\cos4x + 66\cos6x + \cos8x - 65)}{(\cos2x -3)^5}\\ \Rightarrow f^{(4)}(0)= 32\\
+)\quad f^{(5)}(x)= -\dfrac{8\sin2x(3204\cos2x +2140\cos4x +156\cos6x + \cos8x -573)}{(\cos2x -3)^6}\\
\Rightarrow f^{(5)}(0)= 0\\
\text{Ta được:}\\
f(x)= 1 - x^2 + \dfrac{4x^4}{3} + o(x^5)
\end{array}\)
