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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 withsee more
I am going to assume you know what an irrational number is, given that that should have been taught right at the beginning the topic of surds. When I talk about a “rational” number, here I mean “whole number” or “integer”, although a rational number is a number that is not irrational.
There are two types of problem that you could come across when being asked to rationalise the denominator of a fraction:
The first type is when you are asked to rationalise a fraction which has its whole denominator under a square root. For example 1/sqrt(2) or 5/sqrt(8) etc. To rationalise the denominator here, we somehow have to square the denominator to get a whole number. However, we can’t change the value of the fraction! A great way of doing this is therefore by multiplying the fraction with denominator/denominator as you’ll notice that this is just equal to 1: Let’s say we had the fraction n/d (n for numerator, and d for denominator), if we multiply this by d/d, we get nd/d2 (which is still equal to our original fraction). If ‘d’ was irrational, e.g. d=sqrt(2), then d2 would be rational, e.g. (sqrt(2))2=2, and therefore we have successfully rationalised the denominator! For example, we could multiply 1/sqrt(2) by sqrt(2)/sqrt(2) to get sqrt(2)/( sqrt(2))2 = sqrt(2)/2, or 5/sqrt(8) by sqrt(8)/ sqrt(8) to get 5sqrt(8)/(sqrt(8))2 = 5sqrt(8)/8.
The second type is a little trickier as it relies on a concept we call “the difference of two squares”. First, let’s consider (x+y)(x-y) = x2-y2; this result is known as “the difference of two squares”, and it is incredibly useful because if either ‘x’ or ‘y’ were irrational, then we now know that (x+y)(x-y) is rational. For example, if x is irrational (e.g. x=sqrt(2)) and y is rational (e.g. y=3), then (x+y)(x-y)= x2-y2 =(sqrt(2))2-32=2-9=-7, or if x is rational (e.g. x=4) and y is irrational (e.g. y=sqrt(5)), then (x+y)(x-y)= x2-y2=42-(sqrt(5))2=15-5=11. Now let’s return to the second type of problem: you could be asked to rationalise the denominator of a fraction whose denominator comprises of a rational number added to a whole number, for example 1/(1+sqrt(2)) or 3/(sqrt(3)-5) etc. This is definitely more tricky than the first type of problem, but here is where we use what we’ve just learned about “the difference of two squares” to make life a lot easier for us: The denominators of these fractions look a lot like the examples I gave for (x+y) or (x-y) (so 1+sqrt(2) is just (x+y) where x=1 and y=sqrt(2) etc.) so, we probably want to find a way to multiply the denominator of the fraction by the ‘opposite form’ in order to rationalise it. For example, multiplying (1+sqrt(2)) by (1-sqrt(2)) gives us the difference of two squares: 12-(sqrt(2))2=1-2=-1, and multiplying (sqrt(3)-5) by (sqrt(3)+5) = (sqrt(3))2-52=3-25=-22. Now we have a way of turning the denominator into a whole number, we are back to thinking about how we can do this without changing the value of the fraction. Using our trick from the first type, you can see that, if we were given a fraction a/(b+sqrt(c)), then multiplying it by (b-sqrt(c))/ (b-sqrt(c)) wouldn’t change the value ( as (b-sqrt(c))/ (b-sqrt(c))=1) but gives us b2-(sqrt(c))2 on the bottom, which is b2-c, which is a whole number! Going back to my earlier examples, 1/(1+sqrt(2)) x(1-sqrt(2))/ (1-sqrt(2)) = (1-sqrt(2))/-1 = sqrt(2)-1, and 3/(sqrt(3)-5) x(sqrt(3)+5)/ (sqrt(3)+5)=3(sqrt(3)+5)/-22 = (3sqrt(3)+15)/-22 =(-3sqrt(3)-15)/22.see more
I think it’s best I work through an example with you as these problems can vary quite a lot, but the general methods used are the same.
Example: Use the substitution u=x2+5 to find: The integral of (x3/sqrt(x2+5)).dx between the limits of 2 and 1.
So the whole idea of using a substitution here is to simplify the integration for us. The first thing we must do is substitute the given substitution in, otherwise there wouldn’t be much point! In doing this, we also need to replace the .dx in the integration, we do this by finding du/dx and then rearranging for dx; in this particular example, du/dx = 2x, so dx = (1/2x)du, so the integral becomes (x3/sqrt(x2+5)). (1/2x).du =( x2/2sqrt(x2+5)).du. We then use the substitution to get the x’s in terms of u: the numerator, x2 becomes u-5 (as u=x2+5), the denominator, 2sqrt(x2+5) becomes 2sqrt(u). Finally, modifying the limits in terms of u: since the top limit is x=2, then this is equivalent to u=22+5=9, and the bottom limit becomes 12+5=6.
We now have our rephrased integration problem: Integrate ((u-5)/2sqrt(u)).du between the limits of 9 and 6. Notice we can split up the integrand (the thing we’re integrating): =(u/2sqrt(u))-(5/2sqrt(u)). You know that this is equivalent to 0.5u1/2-5.5u-1/2, which when integrated is (1/3)u3/2-5u1/2. The only thing left to do now is apply the limits: [(1/3)(93/2)-5(91/2)]-[ [(1/3)(63/2)-5(61/2)] = 9-15-2sqrt(6)+5sqrt(6) = 7sqrt(6)-6
The steps to solving other substitution problems are very similar to the ones detailed above, obviously the manipulations will be different, but the ideas are the same.see more