Question

how to solve
lim (x-1/2x)/(2x+1/6x)
x→∞

Answer 1

\[ \large \displaystyle \lim_{x \to \infty} \frac{x-\frac{1}{2x}}{2x+\frac{1}{6x}} \] is that your question ?

Answer 2

yes

Answer 3

well the leading terms in the nominator are x and 2x, does that already ring a bell? If not, have you tried factoring out x from both the nominator and the denominator?

Answer 4

i got (1/2x)/{(12x^2+1)/6x}

Answer 5

but i dont think i did it right....

Answer 6

you can do it that way, find the common denominator and then turn out with your algebraic notation, what you think about this?
\[ \large \displaystyle \lim_{x \to \infty} \frac{x-\frac{1}{2x}}{2x+\frac{1}{6x}}=\lim_{x \to \infty} \frac{x \left(1- \frac{2}{x^2} \right)}{x\left( 2+ \frac{1}{6x^2}\right)}=\lim_{x \to \infty} \frac{ \left(1- \frac{2}{x^2} \right)}{\left( 2+ \frac{1}{6x^2}\right)}\]

Answer 7

how did you get \[\frac{ 2 }{ x^2 }\] when you factor out x?

Answer 8

I factored out an \(x\) see the middle expression, you can see that they are equivalent if you distribute the \(x\) inside the parenthesis (...) again

Answer 9

Oh okay! Thank you so much! :D

Answer 10

you're very welcome, when solving such exercises it's also a good idea to get into the idea of 'what is the strongest term in the nominator/denominator', meaning which expressions grows fastest/largest or slowest

Answer 11

I'll work with that idea. Thank you