## Thomas' Calculus 13th Edition

$L(x)=2$
The linearization of $f(x)$ at $x=a$ is given as: $L(x)=f(a)+f'(a)(x-a)$ Knowing that $f(x)=x+\frac{1}{x}$, $f'(x)=1-\frac{1}{x^{2}}$ and $a=1$, we have $L(x)=(1+\frac{1}{1})+(1-\frac{1}{1^{2}})(x-1)=2$