Giải thích các bước giải:
Ta có :
$L=\lim_{x\to 0}\dfrac{e^{2x^2}-\cos3x}{\ln(1-4x^2)+2x^2}$
$\to L=\lim_{x\to 0}\dfrac{e^{2x^2}-(1-2\sin^2\dfrac{3x}2)}{\ln(1-4x^2)+2x^2}$
$\to L=\lim_{x\to 0}\dfrac{e^{2x^2}-1+2\sin^2\dfrac{3x}2}{\ln(1-4x^2)+2x^2}$
$\to L=\lim_{x\to 0}\dfrac{\dfrac{e^{2x^2}-1}{2x^2}+\dfrac{\sin^2\dfrac{3x}2}{x^2}}{-2.\dfrac{\ln(1-4x^2)}{-4x^2}+1}$
$\to L=\lim_{x\to 0}\dfrac{\dfrac{e^{2x^2}-1}{2x^2}+\dfrac 94.(\dfrac{\sin\dfrac{3x}2}{\dfrac 32x})^2}{-2.\dfrac{\ln(1-4x^2)}{-4x^2}+1}$
$\to L=\dfrac{1+\dfrac 94.1}{-2.1+1}$
$\to L=-\dfrac{13}4$
