Question

lim
x→3−

1/ |x − 3|

Answer 1

\[\lim_{x \rightarrow 3}\] but - on top of the 3

Answer 2

so as x approaches 3 from the negative side

Answer 3

https://i.gyazo.com/d06f120f7eb757e34a3f5aa8c607b34f.png is it always going to be positive infinity because of the absolute value?

Answer 4

well if you were to graph this function it would be the normal absolute value function shifted 3 units to the right

Answer 5

oh nm

Answer 6

its 1/|x-3| no just |x-3|

Answer 7

Try utilizing the definition of absolute value in order to redefine the limit.
so having:
\[\lim_{x \rightarrow 3^-}\frac{ 1 }{ \left| x-3 \right| }\]
This will define the limit into two new limits:
(1) \(\lim_{x \rightarrow 3}\frac{ 1 }{ (x-3) } \iff x \ge 3\)
(2)\(\lim_{x \rightarrow 3} \frac{ 1 }{ -(x-3) } \iff x<3\)

Notice, how the lateral limit, this being the "-" above the three x is approximating, implies that we are only interested in those values to the left of x=3, so therefore, we will only focus on limit (2):
\[\lim_{x \rightarrow 3^-}\frac{ 1 }{ -(x-3) }\]
Which is an inmediate limit, which by now you should be able to solve. So I leave it to you.

Answer 8

wait so it's not positive infinity?

Answer 9

Though the limit will tend to -infty, remember that the original function is an absolute value.
It would've been the case if it were to be e^ abs(x-3) but when it is linear, remember that any negativa value is inverted.

Answer 10

why isn't it positive infinity? doesnt the absolute value make it positive?

Answer 11

what's the answer?

Answer 12

For what it's worth, it is positive infinity, and you were right at the beginning.

Answer 13

Yeah, I actually got no time to explain it cause the update of OS came just by.
If we think the situation where it would tend to -infty the limit would still fire to infty cause of the absolute value.

But in reality, the quantity in the denominator \(-(x-3)\) which becomes \(-x +3\) after distributing the minus sign, and the condition being, that "x" approaches "3" with values less than three, which why the minus over the approaching number is negative signifying "from the left".
This means that x could, for instance, have a value of "-2,9999" and adding three to it, it would still become a positive number, which approaches zero.
So we could write \((-x +3) \rightarrow 0^+ \) therefore: \[\lim_{x \rightarrow 3^-} \frac{ 1 }{ -x +3 }=\frac{ 1 }{ 0^+ } =+ \infty\]

Answer 14

Let \(u=|x-3|\), then \(u\ge 0\) for all \(x\)

Hence, as \(x\to 3\) we have \(u\to 0^+\)

thus

\[\Large\lim_{x\to 3}\frac{1}{|x-3|}=\lim_{u\to 0^+}\frac{1}{u}=\infty\]