Question

I have a few precalc problems about finding limits. I'm posting pictures of them so it's clear. Any help would be very appreciated!

Answer 1

|A| itself can be split-ed into 2 regions
|A| = +A when A>= 0
|A| = -A when A<=0
did u know this ?

Answer 2

no, I havent learned anything about this but i just have to get these questions done today

Answer 3

ah, but you have solved limits question before, right ?
and it would be very convenient to help with only one question at a time, please post other after the current question is done

Answer 4

the only time I worked with limits, I was just plugging numbers into a calculator

Answer 5

thats ok.
these are of same types,
lets start with the easier one, the 2nd one

firstly, you know what \(x \to -8^+ \)
and \(x \to -8^-\) mean actually?

Answer 6

heard about left hand and right hand limits?

Answer 7

no :(

Answer 8

ok, then this may take a while, i request you to be patient.

firstly
\(x \to a\) (x approaches a)
mean that 'x' is very near to 'a' but either less than or greater than a

Answer 9

like if i say \(x\to 4\)
it means x can take values like
3.9999 or 3.999999999999 or 4.000001 or 4.00000001
and so on

Answer 10

and it will never make contact with a right?

Answer 11

yes! i am glad u know that :)

so can you now guess what
\(x \to -8^-\)
and
\(x \to -8^+\)
mean ?

Answer 12

x won't equal those values. What is the purpose of the +/- after the number?

Answer 13

thats the catch!
remember that i said when x->4 , x can be greater OR less than 4

now if i say that
\(x \to -8^+\)
means x is very very near to -8, but never = -8, and that "+" indicate, that x can only be GREATER than -8
so, x can take values like -7.999 or -7.999999
makes sense ?

Answer 14

yes

Answer 15

now can you guess what
\(x \to -8^-\)
mean ?

Answer 16

it will only be less than -8

Answer 17

yes,
now coming to limits.
there is a very important result,

lim x-> a f(x) EXIST, only if

\(\large \lim \limits_{x\to a^+}f(x)=\lim \limits_{x\to a^-}f(x)=\lim \limits_{x\to a}f(x)\)

Answer 18

one of the easiest way to find these limits is to just plug in x = a in the function f(x), which you have been doing!

Answer 19

you might have lost me

Answer 20

i just gave you a formula, there was nothing to understand much...or to get lost...

Answer 21

what I got is that I should be plugging in -8 for x

Answer 22

plugging in values does not always work out. and we will consider this later

for the 2nd problem, it works
the only question is to plug in x=-8 in which part of function ? x+9 or -7-x ???

Answer 23

the -7-x

Answer 24

we first need to find
\(\lim \limits_{x\to -8^-}f(x)\)
what will you take for f(x) ???
thats where you think!

well, since \(x\to-8^-\) mean x less than -8
which of these will you take for f(x) ?

x+9 x<-8
-7-x x>=-8

Answer 25

so because we're using (-8)- the value has to be less than -8 and we choose x+9?

Answer 26

absolutely!

Answer 27

similarly for x\(\to -8^+\), since x> -8
we take the f(x) as -7-8, because thats f(x) when x>=-8

Answer 28

knowing that, what is the next step?

Answer 29

using the formula i gave!
\(\large \lim \limits_{x\to a^+}f(x)=\lim \limits_{x\to a^-}f(x)=\lim \limits_{x\to a}f(x)\)

\(\large \lim \limits_{x\to -8^+}(x+9)=\lim \limits_{x\to -8^-}(-7-x)\)

now you can just plug in x =-8 in these!

Answer 30

I should come out with -8, correct?

Answer 31

plug in x =-8 in both of them

Answer 32

oh, so I get 1 as the answer

Answer 33

\(\large \lim \limits_{x\to -8^+}(x+9)=\lim \limits_{x\to -8^-}(-7-x) \\ \large -8+9 - -7+8 \\ \large 1=1\)
now since this is true , our limit x-> -8 DOES EXIST!
and yes, it equals +1 :)

Answer 34

\(*** -8+9 = -7+8\)

Answer 35

first step is to find whether it exist or not

Answer 36

if both of those limits did not have same values, then the original limit itself does not exist!

Answer 37

it does because they are equal?

Answer 38

yess, thats correct
so any kind of doubts for 2nd one ?

Answer 39

Not really (:

Answer 40

cool, lets move on to 3rd :)

can you try it once by yourself ?

for \(x\to -1^-\), which function will you take it as f(x) ?
4-x
5
x+6
?

Answer 41

4-x

Answer 42

correct! and if you plug in x =-1
what do u get?

Answer 43

5

Answer 44

good,
what about \(x\to -1^+\)
which f(x) would u use ?
and after plugging in x =-1
do u still get 5 ?

Answer 45

you use x+6 and it does equal 5 as well

Answer 46

excellent!
so we have
\(\large \lim \limits_{x\to -1^-}f(x) = \lim \limits_{x\to -1^+}f(x)= (\lim \limits_{x\to -1}f(x)= f(-1)) \\ \large 5=5=5 \)
and since thats true, our limit EXIST and it = 5!
any doubts ?

Answer 47

Nope

Answer 48

Is the answer to the first question 2?

Answer 49

lets see :)

now the 1st one

the trick part is the absolute value function.
the general definition is
|X| = X when X>=0
|X| = -X when X<0

Answer 50

so when we have
\(|2-x| \quad x\ge0\)
it is same as saying
\(\large (2-x) \quad \quad (2-x)\ge 0 \\ \large -(2-x) \quad (2-x)<0\)

see if this makes sense, i am just using the definition

Answer 51

it makes sense

Answer 52

and now our f(x) becomes
f(x) = \(\large 10x+2 \quad x < 0 \\\large 2-x \quad x\le -2 \\ \large -2+x \quad x>2 \)
now this becomes similar to your 2nd and 3rs question :)

Answer 53

Is it zero that is equal to x right now?

Answer 54

and since we only need limit at x-> 0,
you would directly plug in x=0 in your f(x) getting the value of limit correctly as 2 :)

Answer 55

Thank you very much!

Answer 56

i hope you understand most of it :)
ans welcome ^_^