Question

find the following limits, or state that they do not exist.
a) \[\lim_{y \rightarrow 2}\frac{ y^4-16 }{ y^4+2y^3-y^2-2y }\]
b)\[\lim_{t \rightarrow -2^+}\sqrt[4]{8+t^3}\]
c) \[\lim_{x \rightarrow 1^+}\left| x-1 \right|/ x-1\]

Answer 1

factor and see if any bad stuff cancels out

Answer 2

actually there are no bad factors, you can directly plugin y=2 into the expression and evaluate

Answer 3

\[\lim_{y \rightarrow 2} (y-2)(y+2)(y^2+4) / y(y^3+2y^2-y-2) \]
I'm stuck here.

Answer 4

the bad stuff lol..

Answer 5

oh, there's no bad stuff... how you know that?

Answer 6

good question :) you will get to know only after plugging in y=2

it is "bad" if you get 0 on both numerator and denominator

Answer 7

because \(\large \dfrac{0}{0}\) is one of the indetermnate forms..

Answer 8

oh yeah! forgot about that. So for a) is 0?

Answer 9

looks good!

Answer 10

is b) also 0? but why is there t -->-2^+?

Answer 11

that means you're walking on x axis from right of -2

Answer 12

\[\lim_{t \rightarrow -2^+}\sqrt[4]{8+t^3} =\sqrt[4]{\lim_{t \rightarrow -2^+}8+t^3} \]

Answer 13

love your drawing :)

Answer 14

plugin t = -2 and evaluate
\[8 + (-2)^3 = 8-8 = 0\]

so we get
\[\lim_{t \rightarrow -2^+}\sqrt[4]{8+t^3} =\sqrt[4]{\lim_{t \rightarrow -2^+}8+t^3} = \sqrt[4]{0} = 0\]

Answer 15

<3

Answer 16

oh i see..

Answer 17

I'm sure how to do c)
Forgot the definition of absolute value.

Answer 18

Easy..
absolute value can never be negative, so we define it in the best way we can think of :

|x| = x if x >= 0
|x| = -x if x < 0

Answer 19

tell me this
is `x-1` positive when x is greater than 1 ?

Answer 20

yes