Use the properties of limits to help decide whether each...

Use the properties of limits to help decide whether each limit exists. If a limit exists, find its value. $$ \lim_{x\to\infty} \frac{4x^2 + 2x}{4x^2 - 3x + 1} $$ Select the correct choice below and, if necessary, fill in the answer box within your choice. A. $$ \lim_{x\to\infty} \frac{4x^2 + 2x}{4x^2 - 3x + 1} = \Box $$ (Simplify your answer. Type an integer or a fraction.) B. The limit does not exist and is neither $$ \infty $$ nor $$ -\infty $$.

Transcript

00:02 Limit is xx squared minus x plus 1 over 4x squared minus 3x minus 1 and we're solving for the limit is x squared infinity.

00:11 Okay, so the first thing i know is that the goal of a limit, okay, the goal of a limit is to eliminate the x's as easily as possible so that you can turn the function into something where you are not evaluating two limits to infinity.

00:27 Because when you limit a value to infinity, for example, 2x, the limit of x approach infinity of 2x by itself is going to be infinity because as x gets larger and larger, it's just going to approach an infinitely large number.

00:43 So what we're going to do is the first step is we're going to try to eliminate the 2x squared and the 4x squared to make them, for example, 2 and 4.

00:52 So let's do that.

00:53 So what i'm going to do is i'm going to take out an x squared over the top and out of the bottom.

00:57 Notice this does not fundamentally change the value of the function because as we know when we take out a 1, it just become the same function.

01:06 So it's going to be x squared over x squared times 2 minus 1 over x because x squared times 1 over x equals negative x plus 1 over x squared.

01:19 Okay.

01:21 And then what we're going to do is we're going to do over 4 minus 3 over x minus 3 over x minus 3 over x, minus 1 over x squared okay so now we can go ahead and evaluate a simpler limit so what we're going to do is we're going to cross out the x squared because i know that if you're taking the limit as x approaching infinity of one is just going to be one so it's going to be the limit as x approaches infinity of this function right here so let's go ahead and do this so when i evaluate infinity for this here and this, these become zero because negative 1 over infinity is just going to become infinitely smaller and is going to be equal to 0.

02:07 And the same thing applies for positive infinity.

02:11 So those are essentially equal to zero when you're solving for limits...

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