When doing one of these limits, you can ignore everything except the highest order term in both the numerator and the denominator. In this case, we care about 2x/x. We can ignore everything else because the highest order term increases at a faster rate than any of the others (think about the graphs of y=x and y=x^2).
As a result, we can compare the two highest-order terms to make a final determination. If the numerator has a higher order than the denominator, the limit will approach infinity as x approaches infinity. If the denominator is larger, the limit will approach 0.
However, in a case like this where the orders are the same, the limit will approach the quotient of the coefficients of the highest order terms. In other words, as x approaches infinity of this particular problem, the limit will approach 2x/x, or 2.
easiest way for this question, just plug infinity in for x and look at what you get.... 2 * infinity is infinity.... 1/ infinity is infinitely small ( zero). you get (infinity) + 0 -5. 5 is so small in comparison to infinity that you are left with infinity. as x approaches infinity the equation approaches infinity.
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When doing one of these limits, you can ignore everything except the highest order term in both the numerator and the denominator. In this case, we care about 2x/x. We can ignore everything else because the highest order term increases at a faster rate than any of the others (think about the graphs of y=x and y=x^2).
As a result, we can compare the two highest-order terms to make a final determination. If the numerator has a higher order than the denominator, the limit will approach infinity as x approaches infinity. If the denominator is larger, the limit will approach 0.
However, in a case like this where the orders are the same, the limit will approach the quotient of the coefficients of the highest order terms. In other words, as x approaches infinity of this particular problem, the limit will approach 2x/x, or 2.
I hope this helped!
Is that supposed to be (2x+1)/(x-5), and not something like 2x + (1 / (x-5)), etc? If so...
Divide the top and bottom by x. This gives you (2 + 1/x) / (1 - 5/x). Now when you take x to infinity, you get (2 + 0) / (1 - 0). So the limit is 2.
Infinity
easiest way for this question, just plug infinity in for x and look at what you get.... 2 * infinity is infinity.... 1/ infinity is infinitely small ( zero). you get (infinity) + 0 -5. 5 is so small in comparison to infinity that you are left with infinity. as x approaches infinity the equation approaches infinity.
Divide all of it by x and look at that limit. They should be equal.