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Greetings Mathematicians in the house. I have been having this difficulties in the problem below. Please help me out!

Discontinuity and continuity of a function.
X^2-81/x-9, (the square of X, minus 81, all over X-9) show that it is a removable continuous function.

Use quotient rule

L'hopital rule will do it

hammmedb:
Use quotient rule

Thanks for the suggestion bro, buh the problem is about continuity of a function, not the derivative.

Gydion:
L'hopital rule will do it

How bro? Please help me do it.

L'hopital rule wont work since we are not given any limit......firstly what is continuity of a function?

To the poster of the question you did not give us limit to work with,provided the limit is given just subtitute it into the function of x given,if the function of x is equal to 1 the function is said to be continous but if the function of x does not equal to 1 it is said to be discontinous function

Snap the question and lemme solve it

Write down the question on a sheet of paper and snap then attach

Your question is not complete.. Snap it .. Also provide the limit.

NDSMELODY:
To the poster of the question you did not give us limit to work with,provided the limit is given just subtitute it into the function of x given,if the function of x is equal to 1 the function is said to be continous but if the function of x does not equal to 1 it is said to be discontinous function

From the question x approaches 9. Thanks for the correction. If I equate 9 for X it will be zero and discontinued. But they say show that it is a removable and a continuous function, thank.

Efewestern:
Your question is not complete.. Snap it .. Also provide the limit.

Limits as X~>9

PigMeat:
Limits as X~>9

okay, Let's do some mathematics..

You have X^2-81/ X - 9 = X^2 - 9^2/ X - 9

= (X - 9) (X + 9) / (X - 9)...... (1)

= from eqn 1, (X-9) will stroke out the denominator (X - 9) so we are left with

(X + 9)

At lim x-> 9 X + 9

= 9 + 9 = 18.

, Mathematicians in the house please correct me if am wrong.. we are all learning..

Lim x^2-81. (X-9)(x+9)
-------- = ------------
X - 9 X-9 The x-9 cancels
X--> 9

Lim x --> 9 (9+9) = 18
As x tends to 9 on x - 9 it's discontinuity is on 18

coobboy:
Lim x^2-81. (X-9)(x-9). Lim (x-9)
-------- = ------------ =. X--> infinity =. Infinity As x tends to infinity on x - 9 it's infinite
X - 9 X-9
X--> infinity

How is X infinite... ?

Efewestern:
How is X infinite... ?

It's a limit question as x goes to infinity

Efewestern:
okay, Let's do some mathematics..

You have X^2-81/ X - 9 = X^2 - 9^2/ X - 9

= (X - 9) (X + 9) / (X - 9)...... (1)

= from eqn 1, (X-9) will stroke out the denominator (X - 9) so we are left with

(X + 9)

At lim x-> 9 X + 9

= 9 + 9 = 18.

, Mathematicians in the house please correct me if am wrong.. we are all learning..

So the limit is continued at 18?

coobboy:
It's a limit question as x goes to infinity

X tends to 9 not infinity

PigMeat:
So the limit is continued at 18?

No its discontinous.. When you solved it wat was your answer. ?

coobboy:
Lim x^2-81. (X-9)(x-9) -------- = ------------ X - 9 X-9 The x-9 cancels X--> infinity
infinity = Infinity As x tends to infinity on x - 9 it's infinite

The part B of the question said find the limit if it continous.

God give me brain na make me too fit contribute for dis kain tin. Una weldone oooo.



I
Know
Even
Know
Where
To
Start
From

Efewestern:
No its discontinous.. When you solved it wat was your answer. ?

I have not solved it. It a problem for me.
U mean it is discontinued at 18? So what is the continuity in it? How do u go abt that?

Subtitute 9 in the equation we get 0/0 which is indeterminate ......9^2-81/9-9 = 0/0 now we can use l'hopital rule by differentiating the numerator and denominator of the function then we have 2x-0/1-0 = 2x then subtitute limit as x->9 which gives 18.........so the function f(x) is said to b discontinous since f(x) is not equal to 1.

Thank god for mathematics hmm nw hw dis tin tak help pesin directly

Run outta thread**

PigMeat:
The part B of the question said find the limit if it continous.

Okay that's even better
The answer is 18 ---- lim x-->>9 (x+9) = (9+9) =18

coobboy:
Okay that's even better
The answer is 18 ---- lim x-->>9 (x+9) = (9+9) =18

NDSMELODY:
Subtitute 9 in the equation we get 0/0 which is indeterminate ......9^2-81/9-9 = 0/0 now we can use l'hopital rule by differentiating the numerator and denominator of the function then we have 2x-0/1-0 = 2x then subtitute limit as x->9 which gives 18.........so the function f(x) is said to b discontinous since f(x) is not equal to 1.

Exactly what I solved earlier, So we all arrive at 18 right ?.. meaning its discontinuos ....

OP any more problem ?

Sandydayziz:
Run outta thread**

Godwinfriz:
Thank god for mathematics hmm nw hw dis tin tak help pesin directly

See them art students.. By there fruit we sha know dem

firstclassmumu:
God give me brain na make me too fit contribute for dis kain tin. Una weldone oooo.



I
Know
Even
Know
Where
To
Start
From

Judging from your moniker I was thinking you gonna deliver .. , So which first class are we talking about now?

Efewestern:
See them art students.. By there fruit we sha know dem




Judging from your moniker I was thinking you gonna deliver .. , So which first class are we talking about now?

are you talking to me?

That you all @ cooboy ndsmelody efewestern y'all contributions were really helpful
.
Here is another question
(1) Prove that limx~>0 (sinx/x)=1
(2) Evaluate limx~>0 (x^3+x^2/x^2)

PigMeat:
That you all @ cooboy ndsmelody efewestern y'all contributions were really helpful
.
Here is another question
(1) Prove that limx~>0 (sinx/x)=1
(2) Evaluate limx~>0 (x^3+x^2/x^2)

2) We have lim X-> 0 (X^3 + X^2/ X^2)

= X^3/X^2 + X^2/X^2

From indices

= X^3-2 + X^2-2

= X^1 + X^0

Remember that anything to power zero is 1 , so we have

= X^1 + 1

= lim X-> 0 0^1 + 1

= 0 + 1

= 1.

that's ur final answer.. As for number 1, will solve it when am chance. bye.

Correct me if am wrong oo.. nobody is perfect..

Using l'hopital rule for the two question......1. Sinx/x subtitute x = 0 we get sin 0/0 = 0/0 which is indeterminate now by differentiating both numerator and denominator we get cosx/1 then subtitute x= 0 we get cos0/1 = 1/1 which gives 1


question 2.

Subtitute x = 0 we have 0/0 which
is indeterminate by finding the first derivative we have 3x^2 + 2x/2x subt x = 0 we get 0/0 again then find second derivative we get 6x+2/2 then subt x = 0 we get 2/2 which gives 1.......