Question

Any tricks related to limits ?

Answer 1

\[\lim_{x \rightarrow \infty}\frac{ 6x^2+2x+1 }{ 5x^2-3x+4 }\]The answer in this case is 6/5.

Answer 2

Lets take another example,

\[\lim_{x \rightarrow \infty}\frac{ 7x^9-4x^5+2x-13 }{ -3x^9+x^8-5x^2+2x }\]
So the answer in this case is -7/3

Answer 3

Let me give you guys a question,

\[\lim_{x \rightarrow \infty}\frac{ 2x^2+5x+1 }{ 3x^2-7x+4 }\]

Answer 4

So the answer to this question is 2/3 < Easy as that.

Answer 5

Does anyone of you know any tricks like these ?

Answer 6

Simple idea: As you go to 100000000000000000 all of those little exponents to the 1 and 5 and such are going to be irrelevant. All that matters is that x^9 (or x^2 for the first).

Then you get 7 times a huge number
over -3 time the same huge number

Those huge numbers go away and you are left with just 7/-3.

Answer 7

So the "trick" is that you take the two that have the highest exponent and forget about the rest. Then you take divide the x^n out and you are left with a limit as x approaches infinity of a constant.

Answer 8

Do you know any other tricks ?

Answer 9

If what you get gives you \[\frac{ x ^{n} }{ x ^{n-1}}\]

Or anything that has the top exponent more than the bottom, then the limit goes to infinity.

However, if the bottom is larger than the top, then the function approaches 0.

Answer 10

I know that i said something which is not related to x approaches to infty.

Answer 11

What is your question?

Answer 12

If it is:

lim y-> inf of 2x^2 then the answer is 2x^2 because the y has no effect on the equation.

Answer 13

Any tricks related to limits other than the ones which approach to infty.

Answer 14

\[\lim_{x \rightarrow 3}\frac{ x^2 }{ (x-3)}\]

I'd assume because otherwise you can just sub in the x.

For this one though, you can use a rule called L'Hopital.

Take the derivative of the top and the bottom and then find the limit of the function.

This is d/dx top / d/dx bottom NOT d/dx top/bottom,