What's wrong with this procedure? Consider the problem of evaluating the $$\lim_{x\to 0}{\frac{x^2+x+1}{2x^2-3}},$$ for example. Then the value is $-1/3.$ But suppose for $x\ne 0$ we rewrite the expression as $$\frac{1+\frac 1x+\frac{1}{x^2}}{2-\frac{3}{x^2}}=g(x).$$ Then the original problem is equivalent to evaluating $$\lim_{x\to\infty}{g(x)},$$ which is equal to $1/2,$ as opposed to $-1/3.$
Clearly the problem is with the second method since the rational function is continuous at $x=0.$ My question is, where exactly has a procedural error been made in the second method? Thank you.
 A: The error is in the step "Then the original problem is equivalent to evaluating ...". It is not. To be precise,
$$
\lim_{x\to 0}\frac{x^2+x+1}{2x^2-3}
=\lim_{x\to \color{red}{0}}\frac{1+\frac1x+\frac1{x^2}}{2-\frac3{x^2}}
\color{red}{\ne}
\lim_{x\to \color{red}{\infty}}\frac{1+\frac1x+\frac1{x^2}}{2-\frac3{x^2}}
$$
A: You have made an error of not converting the limit value when changing the function. Have a look at the equation below to get a better understanding of the flaw in your logic.
$ Y = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x^2+x+1}{2x^2-3} = \lim_{x \to 0} \frac{1+1/x+1/x^2}{2-3/x^2} = \lim_{y \to \inf} \frac{1+y+y^2}{2-3y^2} \neq \lim_{y \to 0} \frac{1+y+y^2}{2-3y^2} $
If it's still not clear, comment and I will elaborate.
A: In your second form, although equivalent, you are in fact dividing by $0$ which you absolutely do not want to do! So you cannot accurately compute the limit as $x\rightarrow 0$ of $\frac{k}{x}$, as you have done. If you were calculating an infinite limit, however, this would be ok (as $\frac{k}{x}\rightarrow 0$).
