Limits Help - Calculus

[vale tree]

A Little bit unsure on what to do here, our teacher didn't explain it very well and I would like some help. If someone can do them I can figure out how to do them based on how you do it.

[inner light]

a) as x approaches 4, what value would g(4) and f(4) have?

[vale tree]

g(4) would be at 2 and f(4) would be at 4.2

[inner light]

f(4) is not quite right

one square unit is 0.5

[vale tree]

f(4) would be 5

my bad

[inner light]

yes

thats all you need

Now you need to evaluate the limit

[vale tree]

Am I just Directly Substituting it in?

[inner light]

using this

Yes

you can graphically see that the limit as x approaches 4 clearly exist for both g and f

[vale tree]

I got -2 plugging it all in

[inner light]

seems right

[vale tree]

So is that all I would need to do for a?

[inner light]

yes b) also

[vale tree]

Well for b is it still Direct Substitution?

Because there is an x by 2

Unless that x is just 1 due to the limit going to 1

[inner light]

Still fine

[vale tree]

Okay I got 3/28

[inner light]

2x as x approaches 1 would be 2

[vale tree]

Okay

[inner light]

how

[vale tree]

oh wait

I just realized

[inner light]

f(1) =
g(1) =

[vale tree]

I used the old things

[inner light]

yes

[vale tree]

f(1)=0
g(1)=-1

[inner light]

yes

[vale tree]

15/8

[inner light]

looks good

[vale tree]

Okay

now for question 2

[inner light]

Is similiar

[vale tree]

If it's approaching 1 from the right side

[inner light]

Yes

1+ approach from right

1- approach from left

[vale tree]

Well f(x) isn't approaching 1 from the right

It's approaching from the left, right?

[inner light]

It is

[vale tree]

So would it just be 0

[inner light]

f and g are continuous at x = 1

[vale tree]

and for g(x) it would be infinity

[vale tree]

so both are infinity

[inner light]

We would have something like -1/0

[vale tree]

Ah I see

so would that be considered negative infinity?

[inner light]

Yes

And from the left side what do you think?

[vale tree]

I see, I am starting to understand this now

From the left side

g(x)=-1
f(x)=0

so if a and b are the same

c is also negative infinity

I think

[inner light]

yes

right thinking

[vale tree]

Thank you

now 1 more part

so at x=2

f(x)-g(x) is (3.5)-(0)

[inner light]

Tricky

As you can see

f is not continuous in x = 2

So we have to limits

[vale tree]

True it's disconnected

[inner light]

We have to see if 2- = 2+ limit

LHS = RHL in x = 2 with f - g

[vale tree]

so I have to do

``
1.9
1.09
1.009

2.1
2.01
2.001
``

to test the limit?

[inner light]

you can but dont need

you need to do this

[shell glade]

adonhs

[inner light]

f has two limits depending from which side you approach, right

at x = 2

Ao for continuity those limits should be equal

This is what you need to check

[vale tree]

So how would I go about writing this?

[inner light]

From the right side f(2+) = 3.5 and from the left f(2-) = 1

[vale tree]

and the g(x) would be 0?

[inner light]

yeah

[inner light]

And now you check with this

You just plug in the according values

What is f and g at x = 2 from left and so on

[vale tree]

So for the left (1-0) is just 1

[inner light]

Nice

And from right

[vale tree]

3.5-0=3.5

[inner light]

Nice

So what can you say

[vale tree]

The Limit Does not Exist

So for b since it's still a disconnect does it not exist aswell?

[vale tree]

or do I say the limit does not exist at values of 1 and 3.5

[inner light]

Yes

Limit doesn not exist

Hence is it continuous or not?

[vale tree]

Not continuous

[inner light]

Right

Wat about b)?

[vale tree]

It wouldn't exist aswell

because it's not continuous

[inner light]

Wrong

Did you check the limits in 3b)?

g • f from left

g • f from right

Same approach like in a) but instead minus it's multiplied

[vale tree]

It's the same

both are 0

1(0)=0

3.5(0)=0

So it is continuous

[inner light]

Nice

So a) not continuous b) it is

[vale tree]

Okay

for a is it not continuous at all or only at x=/=1 and x=/=3.5

[inner light]

Hmm

1 and 3.5 are the y values not x

x is 2

So in x = 2 we all know it is not continuous

[vale tree]

Ah I see

[inner light]

For the rest we may assume it is

[vale tree]

That makes a lot of sense

[inner light]

But we would habe to show it somehow

Continuity is basically that you can draw a function withouz letting go once of the pen

[vale tree]

Yes

[inner light]

Just a vague explaination

But to kinda understand it

[vale tree]

I see, this has been a big help in my understanding.

[inner light]

Nice

[vale tree]

Theres one thing unrelated to this post

I'm not sure when to use this law

Is this when a number is multiplied by the y value?

[inner light]

c is like a scalar

For example

You can pull out factors

[shell glade]

adonhs

[inner light]

Like here I pulled out the 2

Since it's a common factor

[vale tree]

I see

[inner light]

This is just good law to have in mind

To "clean up" calculation or so

[vale tree]

I get it

[inner light]

Not required at all times but sometimes useful

[vale tree]

Yea our teacher gave us a sheet of limit laws but we didn't go over all of them

So you assisting me was a big help

[inner light]

I like your profile design

[vale tree]

Thank You

I like yours too

Cat

[inner light]

Thanks

[vale tree]

ofc

[inner light]

I would show one thing if you dont mind

[vale tree]

sure

[shell glade]

adonhs

[vale tree]

Logarithmatics?

[inner light]

Yes

[vale tree]

Yea we need to do those too

with limits

[inner light]

This a pretty good one to keep in mind

We can basically drag im the limit it's that flexible lol

Helps us with cases such as 0^0 or 1^0 or inf^0

[vale tree]

Oh yeah we didn't do those yet

I should learn them though

[inner light]

Oh ok

mb

[vale tree]

Nono I want to learn it

Its good you brought it up because we are doing it on friday

[inner light]

Oh

Basically thats it

[vale tree]

So what is infinity^0

[inner light]

You can transform those terms with e and ln

[inner light]

undetermined

[vale tree]

is that e^lim x-> (Num) when its infinity

[inner light]

When what is infinity

[vale tree]

Like if you have something thats 1/0

and its infinity

is infinity represented by e?

[inner light]

No

e^limit

[vale tree]

Okay

[inner light]

Ah i understand i think

infinity^0 = e^[ln(inf) • 0]

[vale tree]

ah yeah thats what I was asking

Okay that makes sense

[inner light]

No it is not

Depends

There infinities that are faster, bigger or slower

and smaller

Take a look at this

[vale tree]

Okay

[shell glade]

adonhs

[vale tree]

So that would go to infinity fast

[inner light]

It's basically infinity/infinity

It could be 1, infinity or 0

Anything

What do you think is it here

[vale tree]

I think its infinity

[inner light]

Wrong

[vale tree]

If you use DSP

[inner light]

DSP?

[vale tree]

Direct Substitution Property

[inner light]

Even then

[vale tree]

oh wait

e is its own number

[inner light]

You assume infinity / infinity = infinity

[vale tree]

Im not sure what it is

[inner light]

It's 0

[vale tree]

How do you get 0?

[inner light]

Because e^x beats every polynomial

It's an exponential function which gets bigger much faster

So you have a smaller infinity divided by a bigger one

[vale tree]

I see

[inner light]

And the bigger the denominator gets the more it approaches 0

[vale tree]

so E always takes premise

because it's getting bigger faster

[vale tree]

same idea lol

[inner light]

In the beginning x^100 dominates

But in the long run

e^x

[inner light]

Yes

But that comes with every exponential function

You could also do 3^x

[vale tree]

I see

that shows a lot

[inner light]

Just an example

To get a feeling for that

[vale tree]

I always used to think infinity always takes premise but this is new

Similar if e is in the numerator

[inner light]

Nah infinity can get pretty quickly owned haha

[inner light]

Yeah now its reversed

Now the limit is infinity

Because we divide a bigger infinity by a smaller one

[vale tree]

Yes

[vale tree]

haha

I appreciate the help a lot

[inner light]

No problem thank Jesus

[vale tree]

🙏 ✝️

What denomination are you?

[inner light]

orthodox

[vale tree]

Based

I am a Catholic but likely converting to Orthodoxy

[inner light]

I grew up that way but I've neglected

[vale tree]

What ethnicity are you?

[inner light]

The more you grow up the more you realize the world is cursed

[vale tree]

True

[inner light]

And Jesus teaching are just perfect

[vale tree]

Yes

[inner light]

I am Greek

[vale tree]

Based

I am Assyrian/Chaldean

Greek brother

[inner light]

We all brothers regardless nation

[vale tree]

yes

1 Family

[vale tree]

❤️

[vale tree]

Oh one more question sorry

can you explain the Squeeze theorem

[inner light]

I am not that familiar with that

[inner light]

But basically you start off with two inequalties

[vale tree]

something like this

[inner light]

I understand lol

Basically you wanna make a proof

One side is a little less

The other side is a little more

If both end up with the same limit then that what is in between must also approach that limit

You basicslly start off with a true inequality and arrive with operations to your original one to prove the limit

[vale tree]

I see

So you do som Calc work to figure it out

[inner light]

Yeah

From my understanding, only used it once so far

[vale tree]

Yeah we need to use it for some problems

For something like this I think

Not even sure how to do this problem

[inner light]

hmm

sin(pi/x) is tough

[inner light]

Maybe $e^{-1} \leq e^{sin\frac{\pi}{x}} \leq e^1$ works

[shell glade]

adonhs

[inner light]

because the range of any sin(...) is [-1;1]

And I think you can work with that

now

[shell glade]

adonhs

[vale tree]

So how do I actually solve it?

[inner light]

You start with this

and you try to arrive at what you wanna prove