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### How do I find the limit as x-->infinity of (4x^2+5)/(x^2-6)?

Here you will deal with all problems of the same type in a very similar way.

Intuition first tells us that we can’t really find the limit, as both the top (4x2+5), and the bottom (x2-6) parts of the fraction tend to infinity as x does likewise. However, as is sometimes the case, this is the wrong approach as playing with infinities is a dangerous game!

Note that we’re comfortable at guessing the limit of, say 1/x2 as x tends to infinity (if you’re not, then notice that as x gets very big, 1/x gets very small – the bigger x is, the smaller 1/x is, and since x > 0, 1/x must always be larger than 0 as it can’t be negative, so 1/x gets closer and closer to 0 as x increases. This is another way of saying that “1/x tends to 0 as x tends to infinity”. The limit of 1/x2 is calculated in a very similar way). So in order to get the expression into a format that we’re happy dealing with, it is favourable to divide it by x2/x2 so we’re not changing the value of the expression, but this leaves us with (4+(5/x2))/(1-(6/x2)). Notice now that we can compute the definite limit of each ‘part’ of the fraction: (4+(5/x2)) tends to 4 as x tends to infinity (as 5/x2 tends to 0), and (1-(6/x2)) tends to 1 as x tends to infinity (as -6/x2 tends to 0). So we’re left up with 4/1 = 4. So (4x2+5)/(x2-6) tends to 4 as x tends to infinity.

Note that if you divide by the highest power of x in the expression (over itself) , it will whittle down into something that you’re probably more able to deal with

10 months ago

Answered by Andrew, a GCSE Further Mathematics tutor with MyTutor

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