<s>[INST] derive limit to make a function continuous Here's the problem:
Choose the value of k that makes the following 
function continuous at $x = 1$:
$f(x)=\begin{cases}
\frac{-8x^2 + 48x - 40}{x - 1} & x < 1\\
-2x + k &x \geq 1
\end{cases}$
My steps:
$\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow 1} f(x)$
$\frac{-8x^2 + 48x - 40}{x - 1} = -2x + k$
$\frac{-8(x^2 - 6x + 5)}{x - 1} = -2x + k$
$\frac{-8(x - 5)(x - 1)}{x - 1} = -2x + k$
$-8(x - 5) = -2x + k$
$-8x - (-8)\cdot5 = -2x + k$
$-8x + 40 = -2x + k$
$-8x + 2x + 40 = k$ (See note at bottom)
$-6x + 40 = k$
$-6\cdot1 + 40 = k$
$-6 + 40 = k$
$36 = k$
The solution steps:
For f to be continuous at $x = 1$ we need $\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow1} f(x) = f(1)$
First lets evaluate lim x→1− f(x). Since x < 1 as x approaches 1 from the left, 
f(x) = (−8x^2 + 48x − 40)/(x − 1) as x→1−
So lim x→1− f(x) = lim x→1−  (−8x^2 + 48x − 40)/(x − 1)
Start by factoring the numerator. 
In this case we find: 
lim x→1− (−8x^2 + 48x − 40)/(−3x^2 − 15x + 18) = lim x→1− (−8x + 40)(x − 1)/(x − 1)
Aside: can anyone explain (−3x^2 − 15x + 18) in the denominator?
Since we are taking the limit as x→1−, we may assume that x ≠ 1. 
Canceling factors we see: 
\lim x→1− (−8x + 40)(x − 1)/(−3x − 18)(x − 1) = \lim x→1− (−8x + 40)
We can evaluate this linear function at x = 1. 
Hence, 
lim x→1− −8x^2 + 48x − 40)/(−3x^2 − 15x + 18) = −8(1) + 40. 
Thus, lim x→1− f(x) = 32. 
Thus lim x→1− f(x) = 32 
lim x→1+ f(x) = lim x→ (−2x + k), because x > 1 as x approaches 1 from the right. 
lim x→ (−2x + k) = −2⋅1 + k = −2 + k = f(1) because linear functions are continuous. 
For f to be continuous, these limits have to be equal. Thus 32 = −2 + k 
When k = 34, f is continuous.

NOTE: This is where I diverged from the solution by moving -2x to the left. My question is, assuming the given solution is correct, why is this nominally valid algebraic step not permitted?
 [/INST] You solution is correct, but it's an overkill. After you have correctly found limit from the above, set limit from the below $f(1) = -2 \cdot 1 +k = 34$ and solve for $k$. Continuity is ensured by the fact that $1$ is included in the definition of the limit from below. 
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