Compute $\lim_{x \rightarrow 4} \frac{(2x^2 - 7x -4)}{(-x^2 + 8x - 16)}$

Question

I'm trying compute $\lim_{x \rightarrow 4} \frac{(2x^2 - 7x -4)}{(-x^2 + 8x - 16)}$ without L'Hopital's rule.

$\textbf{My attempt:}$

$\lim_{x \rightarrow 4} \frac{(2x^2 - 7x -4)}{(-x^2 + 8x - 16)} = \lim_{x \rightarrow 4} \frac{2(x + \frac{1}{2})(x - 4)}{(-1)(x - 4)^2} = \lim_{x \rightarrow 4} \frac{2(x + \frac{1}{2})}{(-1)(x - 4)}$

I'm stuck here. How can I eliminate $x- 4$ to compute the limit? Thanks in advance!

Answer 1

What you did is correct, and it prove that the limit doesn't exist in $4$, since you get $+\infty $ in $4^-$ and $-\infty $ in $4^+$.

Answer 2

The factors are not correct. \begin{align*} \lim_{x\rightarrow 4^{+}}\dfrac{2x^{2}-7x-4}{-x^{2}+8x-16}&=\lim_{x\rightarrow 4^{+}}\dfrac{(2x+1)(x-4)}{-(x-4)^{2}}\\ &=\lim_{x\rightarrow 4^{+}}\dfrac{2x+1}{-(x-4)}\\ &=-\infty. \end{align*}

And so the limit is $\infty$ for $y\rightarrow 4^{-}$, as @Cameron Williams has pointed out, the limit does not exist even in the extended real sense.

Answer 3

Note that the limit doesn't exist, indeed as an alternative let $x=y+4$ with $y\to 0$ then

$$\lim_{x \rightarrow 4} \frac{(2x^2 - 7x -4)}{(-x^2 + 8x - 16)}=\lim_{y \rightarrow 0} \frac{(2(y+4)^2 - 7(y+4) -4)}{(-(y+4)^2 + 8(y+4) - 16)} =\lim_{y \rightarrow 0} \frac{2y^2+9y}{-y^2}=\lim_{y \rightarrow 0} \frac{2y+9}{-y}$$

thus the limit doesn't exist since $\lim_{y \rightarrow 0^+}\neq\lim_{y \rightarrow 0^-}$.

Answer 4

Your question of the limit

$$\lim _{x\to \:4+}\left(\frac{2x^2-7x-4}{-x^2+8x-16}\right)=+\infty$$ In fact:

$$\lim _{x\to \:4+}\left(\frac{2x^2-7x-4}{-x^2+8x-16}\right)=\lim _{x\to \:4+}\left(-\frac{2x+1}{x-4}\right)=$$ $$=-\lim _{x\to \:4+}\left(2x+1\right)\cdot \lim \:_{x\to \:4+}\left(\frac{1}{x-4}\right)=-9\cdot \infty =-\infty$$ (with numerator/denominator in the 2nd step).

After

$$\lim _{x\to \:4-}\left(\frac{2x^2-7x-4}{-x^2+8x-16}\right)=+\infty$$ With same steps you have

$$-\lim _{x\to \:4-}\left(2x+1\right)=-9$$ and $$\lim _{x\to \:4-}\left(\frac{1}{x-4}\right)=-\infty$$

Hence

$$\lim _{x\to \:4+}\left(\frac{2x^2-7x-4}{-x^2+8x-16}\right)=-9\cdot (-\infty)=+\infty$$