Question

Find the limit: lim as x approaches infinity 1-2x / sqrt(9x^2 +5)

Answer 1

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\) like this?

Answer 2

yes

Answer 3

\(\large\color{slate}{\displaystyle-\lim_{x \rightarrow ~\infty }\frac{(2x-1)}{\sqrt{9x^2+5}}}\)

now apply LHS

Answer 4

I factored the top out of -1 and took the constant (the -1) out.
as lim cf(x) = c lim f(x)

Answer 5

no not yet

Answer 6

oh, then disregard what I said, since you haven't learned the L'Hospital's rule

Answer 7

are you asked to do this in a particular way, oor you can use any method?

I would simply use a numerical approach

Answer 8

(to plug in very large number to see what the limit is going to approach)

Answer 9

or you can go about is intuitively

Answer 10

we can use any method. I plugged in the equation into my calculator and everytime I'd plug in larger number it would get closer to -.66636 or some value around that.

Answer 11

intuitively:

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\)

the top of the limit as x goes to very large numbers is just -2x, and the +1 there becomes even less significant the greater x value we choose.
the bottom of the limit as x goes to very large numbers is just √(9x²) , becuase the +5 at some point (just like +1 on top) becomes insignificant.

So you can write it as: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\) (and it will be an equivalent)

Answer 12

oh, I meant to say, to write it as

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{\sqrt{9x^2}}}\)

Answer 13

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{\sqrt{9x^2}}}\)

simplify the square root

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{3x}}\)

x's cancel, and you get?

Answer 14

-2/3 ?

Answer 15

yup

Answer 16

this limit is -2/3

Answer 17

ohhhh! makes so much sense. I had this problem on my exam..... I got it wrong

Answer 18

thank you though for clarifying this that way I don't make the same mistake. Thank you.

Answer 19

Yw....