Which Limit Law should I refer to in my solution?

[Abuda]

Homework Statement

Evaluate the limit below indicating the appropriate Limit Law(s) implemented.
[tex]\lim_{x\rightarrow 0.5}\frac{2x^2+5x-3}{6x^2-7x+2}[/itex]

2. The attempt at a solution

[tex]\lim_{x\rightarrow 0.5}\frac{2x^2+5x-3}{6x^2-7x+2}=\lim_{x\rightarrow 0.5}\frac{2(x-0.5)(x+3)}{6(x-0.5)(x-(2/3))}=\lim_{x\rightarrow 0.5}\frac{2(x+3)}{6(x-(2/3))}=-7[/itex]

So would I be required to state anything when I can out the (x-0.5) factor?
(PS, I'm doing Real Analysis and have learned about proving limits from first principles but I'm now trying to learn about using shortcuts by referencing theorems.)

 

[ehild]

Homework Statement

Evaluate the limit below indicating the appropriate Limit Law(s) implemented.
$$\lim_{x\rightarrow 0.5}\frac{2x^2+5x-3}{6x^2-7x+2}$$

2. The attempt at a solution

$$\lim_{x\rightarrow 0.5}\frac{2x^2+5x-3}{6x^2-7x+2}=\lim_{x\rightarrow 0.5}\frac{2(x-0.5)(x+3)}{6(x-0.5)(x-(2/3))}=\lim_{x\rightarrow 0.5}\frac{2(x+3)}{6(x-(2/3))}=-7$$

So would I be required to state anything when I can out the (x-0.5) factor?
(PS, I'm doing Real Analysis and have learned about proving limits from first principles but I'm now trying to learn about using shortcuts by referencing theorems.)

Did you mean so? Do not mix tex and itex.

Say that x--->0.5 means that x tends to 0.5 but never equals to it, so x-0.5 can not equal to zero so you can divide with it. After that explain that the limit of a sum is the sum of the limits, limit of a product with a constant is also equal to the constant times the limit, and the limit of the fraction is equal to the limit of the numerator divided by the (non-zero) limit of the denominator .

ehild

 

[Abuda]

Thank you very much for helping me and fixing up my latex skills. Your explanation about canceling sounds good to me and the other notes about the algebra of limits.
Alex