prove that $\lim_{x \to 1^{-}}\frac{(x+2)}{2x^{2}-3x+1} = -\infty$

Question

Prove that $\lim_{x \to 1^{-}}\frac{(x+2)}{2x^{2}-3x+1} = -\infty$

Without loss of generality, assume tha $M<0$ and $\delta>0$,

choose $0<\delta<\frac{1}{2}$ such that $1-\delta<x<1$ implies $\frac{3}{M}<{2x^{2}-3x+1}<0$ for that is $\frac{M}{3}>\frac{1}{2x^{2}-3x+1}$

since $\frac{1}{2}<x<1$ also implies $\frac{5}{2}<x+2<3$ it follows that $f(x)<M$ for all $1-\delta<x<1$

I don't know whether this is correct or not if there is any easy method to prove it using definition show it. Also tell me is this correct?

Answer 1

There is definitely an easier way, without going into th rigourous definition of a limit. Try factorising the denominator. Then, substiute x=1-h in the eqn, with h tending to zero. Manipulate the eqn a bit, and you would get your desired result.

Answer 2

$$\frac{x+2}{2x^2-3x+1} = \frac{x+2}{(x-1)(2x-1)} = \frac{1}{2(x-1)} \frac{2x+4}{2x-1}= \frac{1}{2(x-1)} \cdot \left(1 + \frac{5}{2x-1}\right)$$

If $x>\frac 12$ then $1 + \frac{5}{2x-1}> 1$.

Therefore if $|x-1| < \delta: = \min\left(\frac 12,\frac{1}{2|M|}\right)$ then $\frac{x+2}{2x^2-3x+1}<-|M|$