
linear approximation
verify the given linear approximation at a=0, then determine the values of x for which the linear approximation is accurate to within 0.1
$\displaystyle \sqrt{1+x}\approx1+\frac{1}{2}x$
which assuming means
$\displaystyle \sqrt{1+x}0.1<1+\frac{1}{2}x<\sqrt{1+x}+0.1$
but couldn't get the given answer of
$\displaystyle 0.69<x<1.09$

First, we notice that the domain is $\displaystyle x \ge 1$.
$\displaystyle 1+\frac{x}{2}<\sqrt{1+x}+0.1$
$\displaystyle \implies~ 0.9+0.5x<\sqrt{1+x}$
Since both sides of the inequality are always nonnegative over the domain $\displaystyle x \ge 1$:
$\displaystyle \implies~ (0.9+0.5x)^2<1+x$
$\displaystyle \implies~ 0.81+0.9x+0.25x^2<1+x$
$\displaystyle \implies~ 0.25x^2  0.1x  0.19<0$
Solving this gives $\displaystyle 0.694427<x<1.09443$ which matches the solution you gave.
By the way, the other side doesn't actually give any restriction:
$\displaystyle \sqrt{1+x}0.1<1+\frac{x}{2}$
$\displaystyle \implies~ \sqrt{1+x}<1.1+0.5x$
$\displaystyle \implies~ 1+x < (1.1+0.5x)^2$
$\displaystyle \implies~ 1+x < 1.21 + 1.1x + 0.25x^2$
$\displaystyle \implies~ 0 < 0.25x^2 + 0.1x + 0.21$
Which is always true, so there is no additional restriction here.