I'm supposed to use Lagrange Remainder Theorem to prove that
$$1 + \frac{x}{2} – \frac{x^2}{8} < \sqrt{1+x} < 1 + \frac{x}{2} \text{ } \text{ if } x>0$$
Obviously, the left and right hand sides look like Taylor series, but how would you do this using Lagrange, it isn't obviously intuitive. Thanks!
Best Answer
For the RHS inequality you can even use Mean Value Theorem (which, of course, is a special case of Taylor theorem - where the reminder term contains first derivative).
Let $f(x) = \sqrt{1+x}$. Then the theorem gives that there exists some $\theta \in (0,x)$ such that $$ \frac{f(x)-f(0)}{x-0} = \frac{\sqrt{1+x} - 1}{x} = f'(\theta) = \frac{1}{2 \sqrt{1+\theta}}. $$ Hence $\sqrt{1+x} = 1 + \frac{x}{2 \sqrt{1+\theta}} < 1 + \frac{x}{2}$ since $x > 0$. Observe that, of course, $\frac{x}{2 \sqrt{1+\theta}}$ is the reminder term in the Taylor theorem.
For the LHS by Taylor theorem you get that there exists $\theta \in (0,x)$ such that $$ \sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} (1 + \theta)^{-5/2} x^3. $$ The reminder term is positive for $x > 0$, which proves the inequality.