Answer to Question #264668 in Real Analysis for Nikhil rawat

Question

Using ∈-δ arguments, prove that

Lim (x^3+1) = 9/8

x→1/2

Answer 1

Given $\epsilon$ > 0, we have to find $\delta>0$ such that

if $0< |x-\frac{1}{2}|< \delta$ then | x³+1$-\frac{9}{8}|$ |<$\epsilon$

Now | x³+1 $-\frac{9}{8}|| = |x³-\frac{1}{8}|$

= |(x $-\frac{1}{2})(x²+\frac{1}{2}x+\frac{1}{4})|$

If |x$-\frac{1}{2}|<\frac{1}{2}$ that means $-\frac{1}{2} <x-\frac{1}{2} < \frac{1}{2}$ then

0 < x < 1 <=> $x²+\frac{1}{2}x +\frac{1}{4} < 1²+ \frac{1}{2} +\frac{1}{4}= \frac{7}{4}$

Therefore |x³+1$-\frac{9}{8}| = |x- \frac{1}{2}|$ (x²+$-\frac{1}{2})(x²+\frac{1}{2}x+\frac{1}{4})< \frac{7}{4}|x-\frac{1}{2}|$

Let us choose $\delta =$ min { $\frac{1}{2} , \frac{4\epsilon}{7}$ }

then $0< |x-\frac{1}{2}|< \delta$ => |x³+1$-\frac{9}{8}|| < \frac{4\epsilon}{7}.\frac{7}{4}=\epsilon$

Thus by the $\delta-\epsilon$ method

$lim_{x->\frac{1}{2}}(x³+1) = \frac{9}{8}$