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Finding $$\lim_{x\to\infty} \frac{(2x-5)^4}{(2x^2+1)(3x^2-2)}$$

Do I multiply top & bottom by $\frac{1}{x^2}$ or $\frac{1}{x^4}$? How do I distribute them into the numerator tho?

In the answer given, I think with some typos:

$$... = \lim_{x\to\infty} \frac{(2-5/x)^\color{red}2}{(2+1/x^2)(3-2/x^2)} = \frac{2^{\color{red}4}}{2\cdot 3} = \frac{8}{3}$$

Either way, how do I distribute the $\frac{1}{x^?}$ into the numerator?

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Could you not just run l'Hôpital's Rule a bunch of times? Or do you not have that theorem at your disposal? –  heat death Apr 15 '12 at 3:45
    
You could be rather primitive (like me) and do a term by term application of the limit after expanding the powers. –  000 Apr 15 '12 at 3:50
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The highest power that appears in the numerator and denominator is $x^4$, so you use $\frac{1}{x^4}$. –  Arturo Magidin Apr 15 '12 at 3:56
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1 Answer

up vote 2 down vote accepted

The dominant term is $x^4$ (Imagine that you expanded both the numerator and denominator. What would the highest power of $x$ be?). So, you could either multiply numerator and denominator by $1/x^4$ (see below), or factor as follows: $$\eqalign{ {(2x-5)^4 \over (2x^2+1)(3x^2-2)} &={ \bigl(x(2-{5\over x})\bigr)^4\over x^2 (2+{1\over x^2}) \cdot x^2(3-{2\over x^2 } ) }\cr &={x^4 (2-{5\over x})^4 \over x^2(2+{1\over x^2})\cdot x^2(3-{2\over x^2 })}\cr &={(2-{5\over x})^4\over (2+{1\over x^2})(3-{2\over x^2})}. } $$

The limit as $x$ tends to infinity is $$ {2^4\over 2\cdot 3}=16/6 =8/3.$$

The $(2-5/x)^2$ is probably a typo..


Using the other approach, where you multiply numerator and denominator by $1/x^4$:

In the numerator, to distribute $1/x^4$ over $(2x-5)^4$: $$\textstyle {1\over x^4}(2x-5)^4 = \bigl( {1\over x} (2x-5) \bigr)^4=(2-{5\over x})^4. $$ (just using $a^nb^n=(ab)^n$ here).

In the denominator, you could do the following: $$\textstyle {1\over x^4}(2x^2+1)(3x^2-2) ={1\over x^2}(2x^2+1) {1\over x^2}(3x^2-2)= (2+{1\over x^2})(3-{2\over x^2}) $$ (or just expand $(2x^2+1)(3x^2-2)$ first and then distribute the $1/x^4$ across).

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