- #1

STAii

- 333

- 1

Greetings.

Let me first show how my teacher solves a limit.

For example, take :

Lim(x->6) (x+5)/(x-3)

He takes the number that x is heading to (forgive me for the terminology), which is 6 in our case.

Now put 6 in the place of x in the function after the limit, you get :

6+5/(6-3) = 11/3

Now, there are 4 conditions for the ratio that u will get :

A: NonZero/NonZero : in this case there is no problem

B: Zero/NonZero : no problem in this case too

C: NonZero/Zero : in this case the limit will either be +[oo] or -[oo]

D: Zero/Zero : in this case we have to use some tricks to make the limit back to the case of A, B or C.

Everything i just said was what my teachers says.

Today in class i came up with a question that (the way i see it) does not work on the rules of my teacher. Here it is :

Define :

Now, try to find the limit :

Lim(x->1) (f(x)-1)/(x-1)

If you use my teacher's way, you will reach 1/0, which means that the limit either equals +[oo] or -[oo]. But if you solve the limit, you will find that it equals 1 (unless i am missing sth).

When i told my teacher about this, he said : "no no no no no (maybe 30 times), it is impossible", i told him "just see this question, and either tell me where i went wrong, or ur rule isn't right", he said "this is not MY rule" (then in a silly way) "if you want try to invent ur own theory then call it on your name", i tried to explain my way of solving the problem, but he wouldn't hear me.

Anyway, here is my way :

Take the limit of the numerator, and the limit of the denominator, then write both limits as a ratio, then apply my teacher's rules (A,B,C and D).

Please tell me what do u think. Feel free to ask any questions.

Thanks.

Let me first show how my teacher solves a limit.

For example, take :

Lim(x->6) (x+5)/(x-3)

He takes the number that x is heading to (forgive me for the terminology), which is 6 in our case.

Now put 6 in the place of x in the function after the limit, you get :

6+5/(6-3) = 11/3

Now, there are 4 conditions for the ratio that u will get :

A: NonZero/NonZero : in this case there is no problem

B: Zero/NonZero : no problem in this case too

C: NonZero/Zero : in this case the limit will either be +[oo] or -[oo]

D: Zero/Zero : in this case we have to use some tricks to make the limit back to the case of A, B or C.

Everything i just said was what my teachers says.

Today in class i came up with a question that (the way i see it) does not work on the rules of my teacher. Here it is :

Define :

Code:

```
f(x) = 2 , (x=1)
x , (x[x=]1)
```

Lim(x->1) (f(x)-1)/(x-1)

If you use my teacher's way, you will reach 1/0, which means that the limit either equals +[oo] or -[oo]. But if you solve the limit, you will find that it equals 1 (unless i am missing sth).

When i told my teacher about this, he said : "no no no no no (maybe 30 times), it is impossible", i told him "just see this question, and either tell me where i went wrong, or ur rule isn't right", he said "this is not MY rule" (then in a silly way) "if you want try to invent ur own theory then call it on your name", i tried to explain my way of solving the problem, but he wouldn't hear me.

Anyway, here is my way :

Take the limit of the numerator, and the limit of the denominator, then write both limits as a ratio, then apply my teacher's rules (A,B,C and D).

Please tell me what do u think. Feel free to ask any questions.

Thanks.

Last edited: