$$\lim_{x \to \infty}\frac{x^2+3}{x^3+1} \;\;\;\;\;\;\;\; \textrm{or}\;\;\;\;\;\;\;\; \lim_{x \to \infty}\frac{10x^3}{2x^3+3x^2+6x}$$
Why do we proceed by first dividing each term in the numerator and denominator by the highest power of $x$ in the denominator?
In the first function, we would divide each term by $x^3$, and in the second function, we would divide each term by $x^3$.
Why divide by $x$ with the highest power in the denominator when determining the limit?
Thank you.
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$\begingroup$Because terms of the form $\frac{1}{x^n}$ converge to zero as $x \to \infty$.
In your example above, $ \lim_{x \to \infty}\frac{10x^3}{2x^3+3x^2+6x} = \lim_{x \to \infty}\frac{10}{2+3\frac{1}{x}+6\frac{1}{x^2}} = \frac{10}{2} = 5$.
$\endgroup$ $\begingroup$The rules describing the relation of limits with the arithmetic operations fail when we have the difference, or the quotient of to functions that tend to (+)infinity. But when their limits are zero then it only fails for the quotient. If an expression is only formed by additions (and subtractions) and quotients. By turning parts of this expression from tending to infinity to tending to zero, you are improving the chances of the rules to work. Moreover, by dividing by one of the terms already in the expression you ensure you are no going to run into the quotient of two functions that tend to zero.
The technique then only requires to have an expression that is formed by adding, subtracting, multiplying and dividing terms that tend to infinity. After writing the expression as one quotient of two sums of terms that tend to infinity; if the term that tends to infinity faster can be identified, then the technique will work.
For example: $\lim_{x\rightarrow\infty}\frac{2^x+3^x}{e^x-x}$ can be computed using this same technique.
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