How can i answer a problem in Limit?

Answer 1

using factorization,rationalization, l'hospitals rule, using neighbourhood concept, using standard formulaes and their combinations and finally using your brain.
Happy limit solving,
be cool.

Answer 2

Ok.. u never asked the peoblem So.. nO one will give u the answer what u need..So to get better answer your question should be specific..u know there is No problem Withought Limit .. In calculus u have to check the limit at everystpe of every problem..Anyway
Take any problem.. there are some limits to the problem
like u can take any problem in daily life it has some limits.. But Some problems have unlimited limit ...means infinite..
So.. To answer a question we have to creat some Limits IF it has Infinite Limit .. according to the problem....
Means .. let's take an example..
If u saw a beautiful girl .. then your thinking in your mind has some limit .. may be that u won't know the limit, But If She meet u , then u will behave with her In Some limits which everyone Like , This means u r cleating some limits, and then solve it generally
WHile solving if u get again infinite limit then u have to redefine it and make it properly...
and see some of the mathematical examples . Understand the concept . Good Luck

Answer 3

When you are asked, for example,

What is the limit of (2x + 3) as x approaches 3?

The solutions is just substitute the value.
2(3) + 3
= 6 + 3
= 9

NOTE: It is written in symbols:
lim x-->3 (2x + 3)

------------------------------------------------
Another...
What is the limit of (x² - 6) as x approaches -2?

lim x--> -2 (x² - 6)

we simply substitute
= (-2)² - 6

= 4 - 6

= -2

--------------------------------------------
So much for the easy part...
but what if you are asked to:

Find the limit of (x² - 9)/(x + 3) as x approaches -3?

lim x --> -3 (x² - 9)/(x + 3)

First we try to substitute
= [(-3)² - 9]/(-3 + 3)

= (9 - 9)/(-3 + 3)

= 0/0

Since 0/0 can be any value, 0/0 is "indeterminate". This means the value of the expression at that x is undefined. We cannot directly substitute x. But that doesn't mean the limit doesn't exist. We try to simplify

lim x--> -3 (x² - 9)/(x + 3)

factor (special products)
= lim x--> -3 (x - 3)(x + 3)/(x + 3)

Cancel
= lim x--> -3 (x - 3)

We can now substitute
= (-3 - 3)

= -6.

therefore, the limit is -6.

There are some harder types of limit problems, but for now I will give you the easy part.
I hope this gives a brief introduction to limits.!!

^_^

Answer 4

Understand the concept.

Limit is when a number approches another number till very close but will never reach the number.

Eg. 5/(6x-5).
In this case, to make this expression valid, 6x cannot be 5, or the value of the denominator will be 0. In other words, x cannot be 5/6.

In a graph, the curve will never touch the line y=5/6. The curve will approach 5/6, like for whatever value given to x, the value may be 0.833333333332 or 0.8333333333333333333331, but will never reach 0.83333333333333333333333333333333333 (or 5/6). The line y=5/6 is known as the asymptote which the curve will never intercept it.

Answer 5

The basic limits 101 is that;
lim f(x) = f(h)
x->h

Usually in limits you are required to evaluate you get a seemingly divergent fraction of functions.
i.e. in the form
lim f(x) / g(x)
x->h
say this is a zero over a zero in that f(h)=g(h) = 0
like lim sin x / x
x->0
Now just notice that the ratio of the derivatives would be the same as the ratio of the functions as both functions reach zero (or infinity possibly, the proof is almost elementary and the result is known as L'Hopital's rule)
i.e lim f(a) / g(a) = 0 / 0 = f'(a) / g'(a) = f^(n) (a) / g^(n) (a)
that is for the previous example sin x / x as x->0 is cos 0 =1

Also note that if the exact term appears in a cancelling out fashion, just go it- no pain
i.e for
lim [x^2 - a^2] / x-a = lim (x-a)(x+a) / (x-a)
x->a x->a
cancelling out we get
lim (x+a) = 2a
x->a
which would be the same case had derivatives been used.
Hope this gives you a general idea!

Answer 6

Generally, you set up the problem in terms of a variable that goes to infility or 0 and then take the limit as the variable goes there.

It would help if you could be a bit more specific ☺


Doug

Answer 7

To answer in limit form, or a problem in limit form? Be more specific... give an example problem.

Answer 8

Most lists tlook something life this

x+3=5
1x+4 = 6

So to solve the list we can take one eqation at a time and solve it. Let's solve the secons qeustion for x

x+4=6

So x=2, Agreed?

Now thel's sustitute waht we know for what we don't. In the frist equation we'll replace x with (2) since x=2. So

1 (2) + 4 = 6 OR
2+4=6

Why don't you rpactice on some that the teacher gave you, post the anwers and we'll double check them for you

Answer 9

A limit is a border or a level You can't go over.

If You have a problem in limit, that means that Your problem is not endless and has borders.

Answer 10

What the hell are you asking? What is "Limit"? Be more specific if you want any type of answer next time.