lim 7x^2+8/2x^2-2x-4 x= infinity?

Answer 1

Sorry. The answer of 22 is wrong. The solution is as follows.

When taking a limit, the best idea is to calculate with a value near the specified limit. If that's not particularly feasible, such as in this case, one can reason a solution without actual calculation.

Here, we have a polynomial of order x^2 divided by another polynomial of order x^2. Since the numerator and denominator are both of the same order, we know the solution will be finite and non-zero. We also know that, at larger values, the x and constant terms are essentially meaningless. So, for large x (i.e., x --> infinity) the fraction becomes

7x^2 / 2x^2.

The x^2 terms cancel out and we have

7/2.

Answer 2

I'm not really sure what you're meaning... a few more parantheses would make it clearer. I'm going to guess on the way you want the order of operations based on the way that would make this problem the most interesting...

lim (7x^2+8)/(2x^2-2x-4)

In these cases, the only thing that matters is the terms with the largest exponent, because that will always dominate over the other terms. Just throw the +8, the -2x, and the -4 out. So that limit will wind up being...

lim (7x^2)/(2x^2) = lim (7)/(2) = 7/2

Answer 3

I suppose 7x^2+8/2x^2-2x-4 = 7*x^2+8/(2*x^2)-2*x-4
lim 7*x^2+8/(2*x^2)-2*x-4 = lim 7*x^2-2*x + lim 8/(2*x^2) - 4
lim 8/(2*x^2) = 0
lim 4 = 4
lim 7*x^2-2*x = lim x * lim 7*x-2 = infinity * infinity = infinity
infinity + 0 - 4 = infinity

Answer 4

What does x approach? Infinity?
So lim (as x->infinity) (7(x^2)+8)/(2(x^2)-2x-4)
You could also graph it.
It's 3 something.
Okay. That is a fraction so you have to see which one is greater.
Ignore everything else but 7x^2 and 2x^2. They cancel out, you're left with 5.

Answer 5

nevertheless i'm particular there is a few extensive algebra you need to apply along with your derivatives, basically think of of it logically. in case you took the shrink it may be infinity over infinity, and indeterminate style. So take a spinoff of the numerator and denominator independently. the spectacular will bring about a relentless, 2, together as the backside would be some very staggering style, yet that style will nonetheless have x's via the spinoff. this means which you've gotten 2 over some variety of infinity providing you with 0 as an answer. because of the fact the exponential style will constantly "develop" quicker than the linear style, you're assured an answer of 0. in case you do no longer have faith me, I also have a TI-89 calculator and checked to confirm i'm spectacular. that's the respond it gave me too. stable luck in Calc! basically FYI, the respond given above is fake because of the fact he by risk entered it so as that the internet web site concept it became (2x-4)/(2x+4) all raised to the (2x+7)th exponent. that's why he got here up with a different answer. in case you do it properly, it is 0, no longer one million/e^eighth.

Answer 6

lim 7x^2+8/2x^2-2x-4
=lim(7+8/x^2)/(2-2/x-4/x^2)
=7+0/2-0-0=7/2

Answer 7

use L'Hopital's rule
differentiating:
14x+8x-2-4
further diff.:
14+8=22
therefore when x approaches infinity the function approaches the value 22

Answer 8

Using L'Hopital's rule the right way:

lim as x-->∞ (14x/4x-2) = still can't tell
lim as x -->∞ (14/4)

7/2 is the answer.

cow's method is also legitimate.