Question

$f(0) = {Lim}_{x \rightarrow 0}\frac{e^{tanx} - e^{x} + \mathcal{l}n(secx + tanx) - x}{tanx - x}$ Applying L'hospital rule

$${\Rightarrow f(0) = {Lim}_{x \rightarrow 0}\frac{e^{tanx} \cdot \sec^{2}x - e^{x} + secx - 1}{\sec^{2}x - 1} }{\Rightarrow f(0) = {Lim}_{x \rightarrow 0}\frac{e^{tanx}\left( \sec^{2}x - 1 \right) + \left( e^{tanx} - e^{x} \right) + secx - 1}{\tan^{2}x} }{\Rightarrow f(0) = {Lim}_{x \rightarrow 0}\left( e^{tanx} + \frac{e^{x}\left( e^{tanx - x} - 1 \right)}{\tan^{2}x} + \frac{1}{secx + 1} \right) }{\Rightarrow f(0) = 1 + 0 + \frac{1}{2} = \frac{3}{2}}$$