Question

Evaluate the following limit:
\[ \lim_{x \rightarrow -\infty} \frac{9x-4}{\sqrt{x^2+5x+3}}\]
Current problem: please refer to my first comment here. Why should we add a negative sign for the denominator?

Answer 1

try dividing both sides by x

Answer 2

Near infinity the numerator is like 9x and the denominator is like x.
How is the ratio look like?

Answer 3

\[
\frac{9x}x=9
\]

Answer 4

\[ \lim_{x \rightarrow -\infty} \frac{9x-4}{\sqrt{x^2+5x+3}}\]\[ =\lim_{x \rightarrow -\infty} \frac{\frac{9x-4}{x}}{\frac{\sqrt{x^2+5x+3}}{-\sqrt{x^2}}}\]Any problems so far?

Answer 5

yep ... one way of looking it
\[ \sqrt{x^2 + ..x + 222 .. } \approx x\]
as x approaches infty

Answer 6

Oops ... -inf ... didn't note.
this hold for both case x-> -inf
\[ \sqrt{x^2 + ..x + 222 .. } \approx x\]

Answer 7

no probs ...so far.

Answer 8

Why?

Answer 9

I put a negative sign in front of \(\sqrt{x^2}\), why no problems for that? In other words, why should I do so?

Answer 10

hold on ... now i'm confused. kinda forgot how to show it.

Answer 11

if you make this approximation \( \sqrt{x^2 + ..x + 222 .. } \approx |x| \) then you will get answer

Answer 12

I'm sorry but I haven't learnt that approximation, would you mind explaining it or..?

Answer 13

let do this , change x = -a, and change the limits a->+inf

Answer 14

I hope you don't mind me asking... But can we really do it in that way, changing the variable and the value it approaches?

Answer 15

yep ... of course we can do ...

Answer 16

this what we call substitution.

Answer 17

Okay. So, a = -x
\[ \lim_{x \rightarrow -\infty} \frac{9x-4}{\sqrt{x^2+5x+3}} = \lim_{a \rightarrow \infty} \frac{-9a-4}{\sqrt{(-a)^2+5(-a)+3}}= \lim_{a \rightarrow \infty} \frac{-9a-4}{\sqrt{a^2-5a+3}}\] Is that correct up to now?

Answer 18

yep.

Answer 19

\[= \lim_{a \rightarrow \infty} \frac{-9a-4}{\sqrt{(-a)^2-5a+3}}\]\[=- \lim_{a \rightarrow \infty} \frac{9a+4}{\sqrt{(-a)^2-5a+3}}\]\[=- \lim_{a \rightarrow \infty} \frac{\frac{9a+4}{a}}{\sqrt{\frac{(-a)^2-5a+3}{a^2}}}\]\[=- \lim_{a \rightarrow \infty} \frac{9+\frac{4}{a}}{\sqrt{1 - \frac{5a}{a^2}+\frac{3}{a^2}}}\]\[=\frac{-9}{\sqrt{1}}\]\[=-9\]Right?

Answer 20

yep ...

Answer 21

Hmm... Interesting...
Would you mind explaining this step: change x = -a, and change the limits a->+inf ?

Answer 22

I'm more interested in the part ''change the limits a-> +inf''. How does it work?

Answer 23

let a = -x
when x->0, a= ?
when x->-inf, a= ?

Answer 24

oh ... that's just clever manipulation i devised. the method is called ... anyhow get answer.

Answer 25

Ahhh, that's clear to me now. Thanks! Should I change the limit to do similar question too?

Answer 26

depending on the question ... and answer. just observe carefuly

Answer 27

And also, what is the usual practice of doing such question? I saw the professor did in the way as I've typed in the first comment, but I'm not quite sure why she added a negative sign there. Her explanation was ''it's because when x-> -inf.''

Answer 28

kinda true ... but changing variable is more cleverer.

Answer 29

there a multiple method to approach problems.

Answer 30

I'm not sure if changing variable is *accepted* (certainly it works this case as the answer matches), and I don't quite understand the professor's way to solve it. I would appreciate if you or anyone can help me understand. This has been a trouble to me for more than a year actually.

Answer 31

look at the graph of both numerator and denominator ...

Answer 32

graph?

Answer 33

the numerator goes to -9 times infinity
the denominator goes to + infinity
---------------------------------
what would the ratio would be like?

Answer 34

Can we really cancel the ''infinity'' ?

Answer 35

hm .. not really.
this doesn't go to infinity ... it rather tends to infinity

Answer 36

that's how you get how a function behaves around some point. i like the concept of limit ... the whole calculus is based on it.

Answer 37

tends to inf., but can we know what is near to infinity? (well, that's probably what we're looking for in limit though..)

Answer 38

yep.

Answer 39

1/x -> 0 as x-> inf
1/x behaves as 0 as x tries to go higher and higher.

Answer 40

I got that concept. Just not sure of the negative sign. It's such a trouble.

Answer 41

well ... try one of those two approaches ...
also try discussing with your prof.

Answer 42

Alright, thanks for your help. I'll just leave the question open to see if anyone can help. Thanks again for your time!

Answer 43

yw