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

1. SolomonZelman

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

2. anonymous

yes

3. SolomonZelman

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

4. SolomonZelman

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

5. anonymous

no not yet

6. SolomonZelman

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

7. SolomonZelman

are you asked to do this in a particular way, oor you can use any method? I would simply use a numerical approach

8. SolomonZelman

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

9. SolomonZelman

or you can go about is intuitively

10. anonymous

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.

11. SolomonZelman

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)

12. SolomonZelman

oh, I meant to say, to write it as $$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{\sqrt{9x^2}}}$$

13. SolomonZelman

$$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{\sqrt{9x^2}}}$$ simplify the square root $$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{3x}}$$ x's cancel, and you get?

14. anonymous

-2/3 ?

15. SolomonZelman

yup

16. SolomonZelman

this limit is -2/3

17. anonymous

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

18. anonymous

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

19. SolomonZelman

Yw....