Why does this limit not exist?

Question

abbles · Answer

limit as x approaches 5
(x^2 - 5x + 6)/(x-5)

I factored (x-3)(x-2)/(x-5)
And I'm hot sure what to do from there... how would I know the limit DNE?

welshfella · Answer

As x approaches 5 from below the denominator becomes a very small negative value so the function approaches a limit of - infinity.
Ans as x  approaches 5 from above  he limit is positive infinity.

welshfella · Answer

So we say as x approaches 5 the limit does not exist.

agent0smith · Answer

I think it's only possible to plug in numbers near x=5 to check, as welsh said.

abbles · Answer

Okay, so I would just plug in a couple numbers and check the values?
How would I have recognized this one as a DNE?  for future problems

agent0smith · Answer

Yes. And remember that a limit DNE when the LHL and RHL are not equal.

abbles · Answer

Okay, thanks agent!

irishboy123 · Answer

if you're looking for some waffle to please teacher, rather than saying you used a calculator, this kinda thing can work, especially the change in the limit

$\lim\limits_{x \to 5} \dfrac{x^2 - 5x + 6}{x-5}$

let $x = 5 + h, |h| <<1$

$= \lim\limits_{h \to 0} \dfrac{(5+h)^2 - 5(5+h) + 6}{h}$

$= \lim\limits_{h \to 0} \dfrac{(h+2)(h+3)}{h}$

for $|h| <<1$, $(h+2)(h+3) > 0$

$\implies sgn ( \lim\limits_{h \to 0} \dfrac{(h+2)(h+3)}{h}) = sgn( h)$

agent0smith · Answer

@IrishBoy123 do you ever learn to quit boring everyone? https://youtu.be/B35vsgBzMh4

abbles · Answer

Wooooah Irish boy, way outta my league D:

irishboy123 · Answer

@agent0smith 

nah mate!

i'll even bore saddos that stand in the toilet all day  😆

welshfella · Answer

lol!