Limit problem

[late radish]

1/4

Need solution @eager forum ???

[eager forum]

Yeah. I remember my prof saying its 1 idk if negative or postiive. She gave answer and not solution.

[late radish]

Take ✓(x+4) = y
i.e. x = y² - 4
(y - 2)/(y² - 4)
1/(y + 2) lim y->2

[stone lake]

multiply numerator and denominator by conjugate

[eager forum]

Could you tell me methods to solve this besides Conjugate and Substitute method? The concept is very easy to understand but the questions are very hard.

I mean is there a sheet how to factor different types of questions? The teacher gave us a simple algebra cheat sheets that doesn't follow the questions she gives.

[placid mesa]

She who

[late radish]

@eager forum bro show her the solution

[coarse quiver]

If the concept is easy, then question applying the concept shouldn't be hard

Also conjugate is the standard algebra trick here

Best to not beat around the bush and just learn it

[eager forum]

alright. thank you!

[frail igloo]

@eager forum has given 1 rep to @coarse quiver

[teal rover]

Notice that this limit is also $\lim_{x\to 0}\frac{f(4+x)-f(4)}x$ with $f\colon x\mapsto\sqrt x$

[willow dune]

Gillian Seed

[teal rover]

So it's just $f'(4)=\frac 1{2\sqrt 4}=\frac 14$.

[willow dune]

Gillian Seed

[late radish]

Bc i take √x+4 = y

So x = y²-4 and numerator become y-2
And y²-4 = (y-2)(y+2)

So your expression will be
(y-2)/(y-2)(y+2)
1/(y+2)

[eager forum]

How did you remove the radical? cancel out the fraction? like Sqrt[x^2-4] ?

I mean the exponent?

[late radish]

I take ✓(x+4) = y

y² - 4 in place of x
But your steps is wrong

[late radish]

Yeah -4 and 4 cancel each other and left with y-2

[eager forum]

oh btw this is different question and thank you for the previous one

[late radish]

The last step is wrong it should be like this
(x-1)(✓x²+3 + 2)/x²+3 - 4
(x-1)(✓x²+3 + 2)/x²-1
(✓x²+3 + 2)/x+1
(✓1+3 + 2)/1+1
(✓4 + 2)/2
(2 + 2)/2
4/2
2

[eager forum]

why is x substituted by zero?

[eager forum]

✓x²/x = ✓x? then substitute zero?

[late radish]

@eager forum

[eager forum]

hello again is this right? I did double sided limits since I don't think it can be evaluated much more.

[late radish]

It is divergent function bro
The answer is infinite

@eager forum

[eager forum]

How do you differentitate indeterminate form to that? I thought since it gives zero with onesided I just try to do two sided limits.

[late radish]

But your two sided limit has flaws

[late radish]

Use L' hopital

[eager forum]

I did the same with this question

[eager forum]

Oh what are those flaws. Is there a criteria for that method to be usable?

[late radish]

You should use h tends to zero and put it in place of x

[late radish]

Divergent function

[eager forum]

I researched about it and came to this solution. Which one do you is right?

[eager forum]

oh the second option I could use L'hospital again to get 2/2 = 1 right?

[eager forum]

Idk how to from here

[late radish]

√{(2/x + 3)/(1 - 1/x)
√{(0 + 3)/(1 - 0)
√3

[late radish]

idk
You didn't give what x tends to

[eager forum]

its this one. I continued on from there

[late radish]

The second one is not right
Using L'hospital it will be 2/2 = 1

[eager forum]

is this right? the first question is discontinuous jump x=1 and removable discontinuous is x=8.

[eager forum]

oh thank you

[frail igloo]

@eager forum has given 1 rep to @late radish

[late radish]

Idk
It's messy
But find
Case 1 lim x-> 1+ , lim x->1- and x=1
Case 2 lim x->4+ , lim x->4- and x=4
If the values are not equal
That means that the function is discontinuous at that value of x

[eager forum]

isn't lim x->1- only applicable to the first function, and lim x->1 and lim x->4- to the second function, and lim x->4+ to the third? Or do you mean I substitue (e.g 0.999, 1.001, 3.999, 4.001)?

[late radish]

substitute x=1
But for x->1- use the function x<1
And for x->1+ use the function x>1
And so on

[eager forum]

Ah I see. How do you redefine the function. The examples I found when redefining the function only gives real numbers only.

I am referring to this sample. In terms of my question, do I just substitute the function of x =>4 with the limit?

[late radish]

Take (x²-4)/(x-2)
we are gonna check for removable discontinuous
x-2=0
x=2
but
(x+2)(x-2)/(x-2)
x+2
put x=2
4
x²-x+2 is also give 4 at x=2
Hence, the function will be redefined as mention in the solution

[eager forum]

Jump Discontinuous at x=1

Removeable discontinuous at x=8

a mistake in x -> 1-

Now I am lost. Idk what it means to indicate if continuous at left/right (It's the first and second function right? but how do I state it) or how to redefine the function.

[late radish]

You wouldn't define if it's not a removable discontinuous but for x=4 you'll define it line this first write (16-x²)/(8-4✓x) for x>4 and 8 for x=4 and ✓x³ -✓x +2 for x<4

[eager forum]

does my answer above shows indication of continuous at the left/right? ty for the redefining.

[late radish]

Nah it doesn't bc all the limit value is not equal

[eager forum]

Isn't the first and second function expands continiuously on left and right? or am I understanding it wrong?

[late radish]

Ping it

[eager forum]

How do I indicate

[eager forum]

^

[late radish]

Yeah 👍

[eager forum]

So I just state which function is continuous from my answer?

[late radish]

x=2 will be continuous

[eager forum]

Where is that? the answer from lim x -> 1+? How

[late radish]

Idk what you're talking about

[eager forum]

I mean where is x=2 from? is it from the lim -> 1-?

[eager forum]

As for this question, is it possible to solve for two sided limits?

[late radish]

Yeah you would

[eager forum]

how? its infinity. Is that what I would substitute?

[late radish]

Yeah and 1/infinity means 0

[eager forum]

The left and the right is infinity? since there is no value that I know that approaches infinity from the left and right.

[late radish]

Idk why you're confused onto this
Look
If I say lim x->1- what I meant is take the function where it's given that x<1 but substitute x=1

[eager forum]

So this could also mean the left and the right? Because I was asked to do two sided limits. But I have no idea how to go with this since it approaches negative infinity or infinity. It would be easier if its 1 so I can just substitute with the nearest decimal (1.0001 or 0.9999)

[late radish]

look
lim x->a+ f(x) = lim h->0 f(a+h)

lim x->a- f(x) = lim h->0 f(a-h)

x=a => f(a)

See there is difference but the value I am substituting is x=a

[eager forum]

what would be the value of h? substituting value nearest to zero?

[eager forum]

Hey man after checking a few videos I now understand what you're getting at and what I'm misunderstanding. thanks