Question

Find the limits of the ff. functions. If the limit does not exist, specify if it is positive or negative infinity. If it's neither, just write DNE.

PLEASE HELP

Answer 1

\[1. \lim_{x \rightarrow +7} \frac{ x ^{2}+4x-3 }{ 4x-28 }\]

Answer 2

oops. \[1. \lim_{x \rightarrow 7^{+}} \frac{ x ^{2}+4x-3 }{ 4x-28}\]

Answer 3

you know whats the meaning of 7+ ??

Answer 4

yup. from the right of 7?

Answer 5

so, 4x-28 will be positive or negative ??

Answer 6

uhm. that where I got confused. when i substistute 7, it'll be undefined then. but with 8, it'll be positive

Answer 7

ok, 7+ is a number very near to 7 but GREATER than 7
so,
4x will be very near to 28 but GREATER than 28
and 4x-28 will be ..... ?

Answer 8

positive? :)

Answer 9

correct! :)

Answer 10

and denominator is obviously positive, right ?

Answer 11

so the answer here is \[+\infty\] ?

Answer 12

so, when x-> 7+
you limit will be (a positive no.)/(number very near to 0, but positive) = + infinity
yes, correct :)

Answer 13

ask if any doubts :)

Answer 14

aww, thankyou so much but i have like 6 more problems here left. can u pls help me along the way? it our take home long test :)

Answer 15

i will try my best to help :)
as long as i am online...

Answer 16

thanks so much.

Answer 17

here it goes \[2. \lim_{x \rightarrow -3} \frac{ 3x ^{2}+4x-15 }{ x ^{3}+27 }\]

Answer 18

just -3 ?

Answer 19

yup

Answer 20

do i need to find first the LHL and RHL and see if they're equal? :)

Answer 21

yesss!
can u find ?

Answer 22

ill try. ill get back to u in a sec. :)

Answer 23

oh, wait no!
you don't need to :P

Answer 24

because the NUMERATOR is factorable!!

Answer 25

can u factor the numerator and denominator ?

Answer 26

aww. yeah sure. so u dont need to what? to substitute and stuff?

Answer 27

you don't need to find LHL and RHL
try to factor

Answer 28

\[\frac{ (3x-5) (x+3) }{ (x+3) (x ^{2}-3x+9) }\]

Answer 29

is that correct? :)

Answer 30

absolutely! :)
so what gets cancelled out ?

Answer 31

x+3 :)

Answer 32

can i just ask? why is LHL & RHL not necessary? i thought we should need to satisfy its left and right limits? cos it not specified in the given, i mean its just -3 :)

Answer 33

when we can solve by directly substituting the value, LHL and RHL is not necessary :)

Answer 34

ohh. i see

Answer 35

so whatever remains, put x=-3 there ,

Answer 36

oh ok. ;)

Answer 37

\[- \frac{14 }{ 27 }\]?

Answer 38

good :) its correct.

Answer 39

oh, okaay. thankyou. so let me get this staright? when will i use the positive over positive thing from just like in #1? :)

Answer 40

when you need LHL and RHL or when you are not sure about the sgns,
like if x->7 and denominator has x-7
or when x->0 and the denominator has x
we are not sure here, because
x->y just means x is very near to y
smaller or greater, don't know

Answer 41

oh okay. thanks. i think here is a hard one \[\lim_{x \rightarrow -5} \frac{ x ^{2}+6x+5 }{ \sqrt{-x-1}-2 }\]

Answer 42

do u know how to rationalize the denominator ?

Answer 43

i think so. yeah. ill try. so first, i need to rationalize the denominator for ..? :)

Answer 44

you want the reason, why we are doing this ?

Answer 45

yeah :)

Answer 46

ok, here i go.
when we have f(x) = (x-3)(x+2) assume,
then if we put x=3, we get a 0, right ??
which means when we get a 0, (x-3) must be a factor of f(x)

similarly here, when i put x=-5 in the denominator, i get a 0!!
so, x-5 MUST be a factor of denominator, i must find it by rationalizing...

Answer 47

ohhhh. thanks agaiin. haha

Answer 48

similarly for numerator, only difference is its already rationalized, just needed to be factored

Answer 49

and i came up with this btw \[- (\sqrt{-x-1}+2) (x+1)\]

Answer 50

wouldn't the denominator be
-x-1 -4 ??

Answer 51

yeah. but its -x-5? nd i put the negative sign outside so i could cancel out x+5?

Answer 52

*and

Answer 53

oh, you already did that!
good girl :)

Answer 54

and yes, thats correct, now you can simply plug in x=-5 :)

Answer 55

Haha. :)

so.. is the answer -16? :)

Answer 56

or +16 ? O.o

Answer 57

oh! haha. yeah. forgot the (-) outside. haha. my bad

Answer 58

thanks for reminding. haha. overlooked it :D

Answer 59

uhm, excuse me for a sec. is my realization right? regarding the LHL & RHL topic.

uhm, "we can only use RHL and LHL if at first, its undefined (by using the direct substitution property at first)?

Answer 60

see usually i first try to find the common factor, and cancel it, like in last proble, it works most of the time and there we need not use LHL and RHL unless specified.

Answer 61

oh. okaay. i get it. :)

Answer 62

but special cases when there is no common factor, and the denominator =0,
then we need it, like when x->3 and denominator =x-3 and we don';t know whether x-3 will be positive or not

Answer 63

ohh :)

Answer 64

next question :)

Answer 65

this kinda makes me confused. haha :)
\[4. \lim_{x \rightarrow \infty} \frac{ x ^{2}-4x ^{3}+4 }{ 2+3x ^{3}-2x }\]

Answer 66

i kinda get it but idk if my answer is right. :D

Answer 67

ok, here is what i do, there are other methods too.....
i substitute x = 1/y
so that when x->infinity , y-> 0

so, change every x to 1/y
the benefit is that, after you simplify, you will just have to put y= 0 :)

Answer 68

aww, our teacher taught us a diff way tho. :)

Answer 69

he said to divide the numerator and denominator by the highest degree of the denominator. is that correct too? :)

Answer 70

did u get the answer as -4/3 ??
if yes, then correct....


yes, that method is short! but not so simple to get

Answer 71

and ofcourse absolutely correct

Answer 72

yah. -4/3 :DDD

Answer 73

here the reason, when x->inf, 1/x -> 0
hence all of these, 1/x^2. 1/x^3, ... and so on will ->0
hence you can just put them as 0

Answer 74

yup. thats exactly what he taught us :)

Answer 75

cool :)
next Q :)

Answer 76

\[5. \lim_{x \rightarrow -\infty} \frac{ \sqrt{x}+x ^{2} }{ x ^{3}-5x ^{2} }\]

Answer 77

idk what going to happen to \[\frac{ \sqrt{x} }{ x ^{3} }\] XD

Answer 78

the power of denominator is simply greater than numerator
so, as x->-infinity, the denominator, grows more quickly than numerator.
so, final limit just = 0 (anything/ very very large = very near to 0)
you will be able to appreciate this with my method of x=1/y too

Answer 79

ohh. yeah. so ill just substitute 1/y to every x then? :)

Answer 80

and for your professor's method
\(\Huge \dfrac{x^m}{x^n}=x^{m-n} \\ \large \dfrac{x^{(1/2)}}{x^3 } = x^{1/2-3} \)
which will also tend to 0 when x->-infinity

Answer 81

ohh.

Answer 82

sooo im going to use ur method then? :D

Answer 83

whichever you are comfortable with...

Answer 84

both will give u final answer =0

Answer 85

oh...

Answer 86

so whats the value for \[\frac{ \sqrt{x} }{ x^{3} }\]

Answer 87

0

Answer 88

lol. because im writing down my solutions :)

thanks

Answer 89

but i thought its only for \[\frac{ 1 }{ x ^{r} }\]? i mean the zero rule thing? :)

Answer 90

yes, its is

Answer 91

but x^1/2 / x^3 can be converted to x^r
\(\Huge \dfrac{x^m}{x^n}=x^{m-n} \\ \large \dfrac{x^{(1/2)}}{x^3 } = x^{1/2-3}\)
where r>1

Answer 92

whats 1/2-3 ?

Answer 93

oh. i see. so the numerator is open for any number and as long as r>1, its zero, right? :D

Answer 94

-2.5? :)

Answer 95

think like this,
if exponent of numerator is LESS than exponent of denominator, then when x->infinity, it = 0

Answer 96

x^-2.5 = 1/x^2.5
exponent >1
so 0

Answer 97

oh.. i get it :D

Answer 98

this is much easier to understand coming from you :D