Vtyjkyuk

Prove that $lim _xto 1f(x)=3$ where $f:(0,infty)to mathbbR$ is given by $f(x)=fracx^3-1x-1$.

I proved by definition of the limit

$$|f(x)-3|=left|fracx^3-1x-1-3right|=|x^2+x-2|=|x-1||x+2|leq |x-1|(|x|+2)$$

how to processed from this

$x^3-1=(x-1)(x^2+x+1)$.

Hence $lim_xto1fracx^3-1x-1=lim_xto1x^2+x+1=1^2+1+1=3$.

Hint:

From

$$x^2+x-2=(x-1)(x+2)$$ you deduce that the expression will tend to zero thanks to the first factor, and in a finite neighborhood, say $[0,2]$ the second factor remains bounded (does not exceed $4$).

Hence,

$$|x^2+x-2|le 4|x-1|.$$

This allows you to find the $epsilon-delta$ combinations.

As an alternative by definition of derivative with $g(x)=x^3implies g'(x)=3x^2$

$$lim _xto 1fracx^3-1x-1=g'(1)=3$$

or by factorization

$$lim _xto 1fracx^3-1x-1=lim _xto 1,frac(x-1)(x^2+x+1)x-1=lim _xto 1,(x^2+x+1)=1+1+1=3$$

Let $varepsilon > 0$. We have

$$left|fracx^3-1x-1 - 3right| = left|fracx^3-3x+2x-1right| = left|x^2+x-2right| = |x-1||x+2| le |x-1|(|x-1| + 3)$$

Solving the quadratic inequation $y^2+3y - varepsilon < 0$ gives $y in leftlangle frac-3-sqrt9+4varepsilon2, frac-3+sqrt9+4varepsilon2rightrangle$.

Therefore if we pick $delta = frac-3+sqrt9+4varepsilon2$, then $|x-1| < delta$ implies

$$left|fracx^3-1x-1 - 3right| le |x-1|^2 + 3|x-1| < varepsilon$$

We conclude $lim_xto 1 fracx^3-1x-1 = 3$.

Limit of a product is product of the limits. Since the limit of $|x-1|$ is $0$ and the limit of $|x|+2$ is 3, the limit of the product is $0$. For an $epsilon- delta$ argument let $epsilon >0$ and choose $delta in (0,1)$ such that $delta <epsilon /4$. Then $|x-1| <delta$ implies $|x-1| <epsilon /4$ and $|x|+2<4$ so the product is less than $(epsilon /4) (4)=epsilon$, i.e. $|f(x)-3|<epsilon$.

By definition of the limit
$$f(x)-3=fracx^3-1x-1-3=x^2+x-2=(x-1)^2+3(x-1)$$
beginalign
|f(x)-3|
&= |(x-1)^2+3(x-1)| \
&= |x-1|^2+3|x-1| \
&<delta^2+3delta \
&= varepsilon
endalign