Đáp án:

\(\begin{array}{l}
31)\quad \lim\limits_{x\to 0}\dfrac{e^{x^2} - \cos x}{\dfrac12x^2}= 3\\
32)\quad \lim\limits_{x\to 0}\dfrac{e^{ax} - e^{bx}}{\sin ax - \sin bx}= 1\\
33)\quad \lim\limits_{x\to 0}\dfrac{\sqrt[3]{\cos x} - \sqrt{\cos x}}{x^2}= \dfrac{1}{12}\\
34)\quad \lim\limits_{x\to 0}\dfrac{\cos mx - \cos nx}{x^2}= \dfrac12(n^2- m^2)
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
31)\quad \lim\limits_{x\to 0}\dfrac{e^{x^2} - \cos x}{\dfrac12x^2}\\
= \lim\limits_{x\to 0}\dfrac{\left(e^{x^2} - 1\right) + (1 - \cos x)}{\dfrac12x^2}\\
= \lim\limits_{x\to 0}\dfrac{x^2 + \dfrac12x^2}{\dfrac12x^2}\\
= 3\\
32)\quad \lim\limits_{x\to 0}\dfrac{e^{ax} - e^{bx}}{\sin ax - \sin bx}\\
= \lim\limits_{x\to 0}\dfrac{\left(e^{ax} - 1\right) - \left(e^{bx} - 1\right)}{\sin ax - \sin bx}\\
= \lim\limits_{x\to 0}\dfrac{ax - bx}{ax - bx}\\
= 1\\
33)\quad \lim\limits_{x\to 0}\dfrac{\sqrt[3]{\cos x} - \sqrt{\cos x}}{x^2}\\
= \lim\limits_{x\to 0}\left(\dfrac{\sqrt[3]{\cos x}-1}{x^2} + \dfrac{1-\sqrt{\cos x}}{x^2}\right)\\
= \lim\limits_{x\to 0}\left[\dfrac{\cos x - 1}{x^2\left(\sqrt[3]{\cos^2x} + \sqrt[3]{\cos x} + 1\right)} + \dfrac{1 - \cos x}{x^2\left(1 + \sqrt{\cos x}\right)}\right]\\
= \lim\limits_{x\to 0}\left[\dfrac{-\dfrac12x^2}{x^2\left(\sqrt[3]{\cos^20} + \sqrt[3]{\cos 0} + 1\right)} + \dfrac{\dfrac12x^2}{x^2\left(1 + \sqrt{\cos 0}\right)}\right]\\
= -\dfrac16 + \dfrac14\\
= \dfrac{1}{12}\\
34)\quad \lim\limits_{x\to 0}\dfrac{\cos mx - \cos nx}{x^2}\\
= \lim\limits_{x\to 0}\dfrac{(1 - \cos nx) - (1-\cos mx)}{x^2}\\
= \lim\limits_{x\to 0}\dfrac{\dfrac12(nx)^2 - \dfrac12(mx)^2}{x^2}\\
= \dfrac12(n^2- m^2)
\end{array}\)