Chapter 3 - Determining Change: Derivatives - 3.7 Activities - Page 244: 3

$$\lim _{x \rightarrow 0}\left[e^{x}-\ln (x+1)\right] =1$$

Given $$\lim _{x \rightarrow 0}\left[e^{x}-\ln (x+1)\right]$$ By using the method of replacement \begin{align*} \lim _{x \rightarrow 0}\left[e^{x}-\ln (x+1)\right]&=\lim _{x \rightarrow 0}\left[e^{0}-\ln (0+1)\right]\\ &= \lim _{x \rightarrow 0}\left[1 \right]\\ &=1 \end{align*}