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# Find the limit or show that it does not exist.$\displaystyle \lim_{x \to \infty}\frac{\sqrt{x + 3x^2}}{4x - 1}$

## $$\frac{\sqrt{3}}{4}$$

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infinity. That means X gets very large and then larger and larger. Uh the numerator is going to be dominated uh In most cases by the term that has the highest power of X. And the same goes where the denominator? The denominator will be dominated by the term that has the highest power of X. So if we look at the numerator, uh three X squared is the term that has the highest power Becks and uh the whole numerator is going to be dominated by this three X squared term because as X gets very very large, X squared gets even larger and then times it by three gets even larger and adding just X to it, it does make it larger, but the dominating term will be the three times X squared because X already being large. Once you squared it gets even larger. And so the dominating term in the numerator is 20 X squared. And then don't forget we are taking the square root of uh the numerator. So we're going to the whole numerator is going to be dominated by the square root of the three X squared term. Now the denominator is for x minus one, the dominating term. Uh In the denominator, the four x minus one expression is going to be the four X as X gets very large. Four times X gets very large, subtracting one from a very large number does not really change it. Um so four X is going to be our dominating term. Uh In the denominator of this function. So the behaviour of this function as X approaches infinity is going to going to be dominated by the three, the square root of three, X squared over four X. Now the square root of three, X squared is uh squared at three times X. Since the square root of x squared is just X. We don't need to write the absolute value of X because X is going to be positive because we're taking the limit as X approaches positive infinity. Uh And then the denominator we had four times X. Now since we're times in the numerator by X and the denominator by X, we can cancel out the excess. So we end up with square to 3/4. And so this expression will approach the value square to 3/4 as X gets very large towards infinity. Um Now, numerically discredited 3/4. If we put it in the calculator, it's approximately equal 2.433. And I did that because we're going to look at a graph of this function and we're going to we're going to see as ex moves uh to the right along the positive X axis on its way to positive infinity. Uh We're going to look at the values of this function and see if it actually does approach uh the limit of square to 3/4 by seeing if the y coordinate is approaching the number of 0.433. And so here is a graph of our function and uh if we go back to the origin, always good to start at the point X equals zero. Um So here's X equals zero. So if we go to just something like X equals five, let's see what the function value is, 50.469. Okay, so not too far from our limit of 0.433, but let's move further down the X axis as X travels to the right towards positive infinity. So when X is only near 30 we can see that the function value is now 300.439, closer to our limit of 0.433. So we can see that dysfunction will approach our limit as X approaches positive infinity. And so the limit of our function As X approaches positive infinity is the square to 3/4.

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