Question

Can you help me solve this limit?
http://screencast.com/t/ad1VNJN2

The correct answer According to the book is supposed to be sqrt(5) but I get is -infinity

I use These RULES:
(1) if numerator's exponent value is greater then limit approaches infinity
(2) if denominator's exponent value is greater then limit approaches zero
(3) if both numerator and denominator's exponent value are same then limit is the coefficient of highest term in numerator divided by coefficient of highest term in denominator

As you can see I used Rules #1 for my approach.

Answer 1

@ajprincess

Answer 2

@goformit100

Answer 3

Yes your approach is correct :)

Answer 4

But why I get wrong answer!! :(

Answer 5

What have you got?

Answer 6

-infinity

Answer 7

Yes What have you got ?

Answer 8

Have you learned L'Hopital's rule?

Answer 9

\[\lim _{x \rightarrow - \infty}\frac{\sqrt{5x^2 -2}}{x+3}\]\[=\lim _{x \rightarrow - \infty}\frac{\sqrt{\frac{5x^2 -2}{x^2}}}{\frac{x+3}{x}}\]\[=\lim _{x \rightarrow - \infty}\frac{\sqrt{5-\frac{2}{x^2}}}{1+\frac{3}{x}}\]

Answer 10

Can you do it from here?

Answer 11

Wait wait wait Callisto, can you solve it using only the rules I posted above?

Answer 12

I need to see where is my mistaken point of view using THOSE specific rules

Answer 13

#3?

Answer 14

Those rules do not work until you mess around with the initial function for a little bit, then you can apply them.

Answer 15

The way i see it is:\[\frac{ \sqrt{5x^2} }{ x }=\frac{ \sqrt{5}x }{ x }=\sqrt{5}/1\]

Answer 16

Is it how Luigi0 says @Callisto ?

Answer 17

I know it's probably not right

Answer 18

I think the limit is not \(\sqrt{5}\). It should be \(-\sqrt{5}\)

Mathematically, luigi's way is not quite right :|

Answer 19

but you said that I should use rule 3# instead of rule 1# @callisto so we somehow have to make exponents equal

Answer 20

Ha, I know it's not right but hey worth a try

Answer 21

\[\sqrt{x^2} = \pm x\]The degree of x in numerator and denominator are both 1. BUT you can't ignore the constant under the square root. :|

Answer 22

if we cant ignore them then what we do?

Answer 23

ok so \[\lim _{x \rightarrow - \infty}\frac{\sqrt{5x^2 -2}}{x+3}\] divide everything by \(x\)
\[lim _{x \rightarrow - \infty}\frac{ \frac{1}{x} \sqrt{5x^2 -2}}{\frac{1}{x}(x+3)}\]
this is to get it in a form we can solve

Answer 24

then \[lim _{x \rightarrow - \infty}\frac{ \sqrt{5 -\frac{2}{x}}}{1+\frac{3}{x}}\]

Answer 25

now apply your rule #2

Answer 26

\[\lim _{x \rightarrow - \infty}\frac{\sqrt{5x^2 -2}}{x+3}\]\[=\lim _{x \rightarrow - \infty}\frac{\frac{\sqrt{5x^2 -2}}{x}}{\frac{x+3}{x}}\]\[=\lim _{x \rightarrow - \infty}\frac{\sqrt{\frac{5x^2 -2}{x^2}}}{\frac{x+3}{x}}\]\[=\lim _{x \rightarrow - \infty}\frac{\sqrt{\frac{5x^2}{x^2} -\frac{2}{x^2}}}{\frac{x}{x}+\frac{3}{x}}\]\[=\lim _{x \rightarrow - \infty}\frac{\sqrt{5 -\frac{2}{x^2}}}{1+\frac{3}{x}}\]Now, apply #2

Answer 27

Btw, the right answer should be \(-\sqrt{5}\)

Answer 28

it's still positive @Callisto

Answer 29

where did you get the -?

Answer 30

No, if x approaches to -ve infty, and the degree of x is odd, we should get a -ve.

Answer 31

If I apply rule #2 the answer automatically gets to 0