Xét $x\in (-\frac{\pi }{4};\frac{\pi }{4})$. Khi đó $sinx+cosx>0$ và $cosx>0$. Gọi $A$ là giới hạn phải tính, thế thì:$A=\mathop {\lim }\limits_{x \to 0}\frac{ln(sinx+cosx)}{x}$
$=\mathop {\lim }\limits_{x \to 0}\frac{ln[cosx(tanx+1)]}{x}$
$=\mathop {\lim }\limits_{x \to 0}\frac{ln(cosx)+ln(tanx+1)}{x}$
$=\mathop {\lim }\limits_{x \to 0}[\frac{ln(1-2sin^2\frac{x}{2})}{x}+\frac{ln(1+tanx)}{x}]$
$=\mathop {\lim }\limits_{x \to 0}[\frac{ln(1-2sin^2\frac{x}{2})}{-2sin^2\frac{x}{2}}.\frac{sin\frac{x}{2}}{\frac{x}{2}}.(-sin\frac{x}{2})+\frac{ln(1+tanx)}{tanx}.\frac{sinx}{x}.\frac{1}{cosx}]$
$=1.1.(-0)+1.1.\frac{1}{1}$
$=1$.