Find the linear approximation of $LaTeX: \displaystyle f(x) = \ln{\left(x + 1 \right)}$ at $LaTeX: \displaystyle a = 0$ and use it to approximate $LaTeX: \displaystyle \ln{1.26}$

Using the formula for the linearization $LaTeX: \displaystyle L(x) = f'(a)(x-a)+f(a)$ gives: $LaTeX: f(x) = \log{\left(x + 1 \right)} \approx L(x) = x$ To approximate $LaTeX: \displaystyle \ln({1.26})$ use $LaTeX: \displaystyle x = 1.26-1 = 0.26$ . $LaTeX: \ln({1.26}) = f(0.26) \approx L(0.26) = 0.26$