Question

Please help me Find the following limits:

1. limit (as x approaches infinity) :(sqrt(2(x^2)+x)-sqrt(2)x)

2. lim (as x approaches infinity) : 10^9(sqrt(2(x^2)+x) - sqrt(2)x)

3. lim(as x approaches neg infinity) : (sqrt(2(x^2) +3))/x

Answer 1

have you tried rationalizing the numerator for #1 ?

Answer 2

yes and then i got super lost

Answer 3

\[\begin{align} \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x


\end{align}\]

Answer 4

is that the first limit ?

Answer 5

yes

Answer 6

\[\begin{align} \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x
& = \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x\times\dfrac{ \sqrt{2x^2+x} + \sqrt{2}x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{\left(\sqrt{2x^2+x}\right)^2 -
\left(\sqrt{2}x\right)^2}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\



\end{align}\]

Answer 7

next divide top and bottom by x

Answer 8

so 1/((sqrt(2x^2+x) /x) +sqrt (2)?

Answer 9

\[\begin{align} \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x
& = \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x\times\dfrac{ \sqrt{2x^2+x} + \sqrt{2}x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{\left(\sqrt{2x^2+x}\right)^2 -
\left(\sqrt{2}x\right)^2}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x \color{red}{/x}}{\sqrt{2x^2+x} \color{red}{/x} +
\sqrt{2}x \color{red}{/x}}\\~\\

\end{align}\]

Answer 10

when u push that x inside radical, it becomes x^2

Answer 11

so sqrt (2x^4+x)?

Answer 12

\[\begin{align} \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x
& = \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x\times\dfrac{ \sqrt{2x^2+x} + \sqrt{2}x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{\left(\sqrt{2x^2+x}\right)^2 -
\left(\sqrt{2}x\right)^2}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x \color{red}{/x}}{\sqrt{2x^2+x} \color{red}{/x} +
\sqrt{2}x \color{red}{/x}}\\~\\


& = \lim\limits_{x\to\infty}
\dfrac{1}{\sqrt{2x^2\color{red}{/x^2}+x\color{red}{/x^2}} + \sqrt{2}}
\end{align}\]

Answer 13

oh, ok,

Answer 14

so 1/(sqrt(2+1/x) + sqrt (2)?

Answer 15

\[\begin{align} \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x
& = \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x\times\dfrac{ \sqrt{2x^2+x} + \sqrt{2}x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{\left(\sqrt{2x^2+x}\right)^2 -
\left(\sqrt{2}x\right)^2}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x \color{red}{/x}}{\sqrt{2x^2+x} \color{red}{/x} +
\sqrt{2}x \color{red}{/x}}\\~\\


& = \lim\limits_{x\to\infty}
\dfrac{1}{\sqrt{2x^2\color{red}{/x^2}+x\color{red}{/x^2}} + \sqrt{2}} \\~\\

& = \lim\limits_{x\to\infty} \dfrac{1}{\sqrt{2+1/\color{red}{x}} + \sqrt{2}}


\end{align}\]

Answer 16

yes take the limit now

Answer 17

as x -> infinity
1/x -> 0

Answer 18

\[\begin{align} \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x
& = \lim\limits_{x\to\infty} \sqrt{2x^2+x} - \sqrt{2}x\times\dfrac{ \sqrt{2x^2+x} + \sqrt{2}x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{\left(\sqrt{2x^2+x}\right)^2 -
\left(\sqrt{2}x\right)^2}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x}{\sqrt{2x^2+x} + \sqrt{2}x}\\~\\

& = \lim\limits_{x\to\infty} \dfrac{x \color{red}{/x}}{\sqrt{2x^2+x} \color{red}{/x} +
\sqrt{2}x \color{red}{/x}}\\~\\


& = \lim\limits_{x\to\infty}
\dfrac{1}{\sqrt{2x^2\color{red}{/x^2}+x\color{red}{/x^2}} + \sqrt{2}} \\~\\

& = \lim\limits_{x\to\infty} \dfrac{1}{\sqrt{2+1/\color{red}{x}} +
\sqrt{2}}\\~\\

& = \dfrac{1}{\sqrt{2+\color{red}{0}} +
\sqrt{2}}\\~\\


\end{align}\]

Answer 19

simplify

Answer 20

so as x approaches infinity the limit would be going to 1/sqrt(2) right? Thanks so much!!

Answer 21

try again

Answer 22

in the bottom there are two radicals right

Answer 23

oh wait 1/(2sqrt(2)

Answer 24

yes!

Answer 25

Thanks that really helped me understand what Im trying to get to

Answer 26

for #2,
simply multiply this by 10^9

Answer 27

so on the second one, when you add the 10^9 out in front wouldnt it just be 10^9(1/(ssqrt2) or am I not thinking about that correctly?

Answer 28

you're right, if the limit exists, :
\[\large \lim\limits_{x\to } Cf(x) =C\lim\limits_{x\to } f(x) \]

Answer 29

great, and on the last one can't you just think that the top is approaching infinity and bottom is approaching neg infinity so the whole thing is approaching -infinity? Or am I way off?

Answer 30

\[ \lim\limits_{x\to\infty}10^9 \left( \sqrt{2x^2+x} - \sqrt{2}x\right) = 10^9 \lim\limits_{x\to\infty} \left( \sqrt{2x^2+x} - \sqrt{2}x\right) = 10^9 \left(\dfrac{1}{2\sqrt{2}}\right)\]

Answer 31

you have good intuition,
but its tricky.. u need to work it using limit properties

Answer 32

you will be in problem if the denominator approaches inifnity at faster rate compared to numerator

Answer 33

then, the aggressive denominator pulls the entire fraction to 0 and the limit also will be 0

Answer 34

intuition is good, but it wont work always with limits

Answer 35

lets divide top and bottom by x and see what we get

Answer 36

Thats what i was just working on :) I got sqrt (2+ 3/x)

Answer 37

no wait, i was wrong.. lets first make a substitution

Answer 38

doesn't 3/x approach 0? and we are left with 2 or am i looking at that wrong?

Answer 39

lets turn negative infinity to positive infinity

Answer 40

ok

Answer 41

substitute x = -y
as x-> -infinity, y -> infinity

right ?

Answer 42

\[\begin{align} \lim\limits_{x\to-\infty} \dfrac{\sqrt{2x^2+3}}{x}

&= \lim\limits_{y\to\infty} \dfrac{\sqrt{2(-y)^2+3}}{(-y)}

\end{align}\]

Answer 43

are you happy with above ?
(you will see why it is needed shortly)

Answer 44

\[\begin{align} \lim\limits_{x\to-\infty} \dfrac{\sqrt{2x^2+3}}{x}

&= \lim\limits_{y\to\infty} \dfrac{\sqrt{2(-y)^2+3}}{(-y)} \\~\\

&= \lim\limits_{y\to\infty} \dfrac{\sqrt{2y^2+3}}{-y} \\~\\


\end{align}\]

Answer 45

So you can change the infinity by taking the opposite of the variable?

Answer 46

i have just substituted x by -y

Answer 47

everywhere

Answer 48

since x is going to -infinity,
y has to go to +infinity

right ?

Answer 49

\[\begin{align} \lim\limits_{x\to-\infty} \dfrac{\sqrt{2x^2+3}}{x}

&= \lim\limits_{y\to\infty} \dfrac{\sqrt{2(-y)^2+3}}{(-y)} \\~\\

&= \lim\limits_{y\to\infty} \dfrac{\sqrt{2y^2+3}}{-y} \\~\\

&= \lim\limits_{y\to\infty} -\sqrt{2+3/y} \\~\\


\end{align}\]

Answer 50

take the limit now

Answer 51

as y -> infinity,
1/y -> 0

so u get :

\[\begin{align} \lim\limits_{x\to-\infty} \dfrac{\sqrt{2x^2+3}}{x}

&= \lim\limits_{y\to\infty} \dfrac{\sqrt{2(-y)^2+3}}{(-y)} \\~\\

&= \lim\limits_{y\to\infty} \dfrac{\sqrt{2y^2+3}}{-y} \\~\\

&= \lim\limits_{y\to\infty} -\sqrt{2+3/y} \\~\\

&= -\sqrt{2+0} \\~\\


\end{align}\]

Answer 52

simplify

Answer 53

so it is -sqrt 2?

Answer 54

yup!

Answer 55

u may use wolfram to double check ur answers in future

http://www.wolframalpha.com/input/?i=+lim%28as+x+approaches+neg+infinity%29+++++%28sqrt%282%28x%5E2%29+%2B3%29%29%2Fx

Answer 56

Thanks, so we added the negative because it was to negative infinity originally?

Answer 57

yes you can use that trick for evaluating limits at neg infinity

Answer 58

convert it to pos inifnity by using that substitution x = -y and work it

Answer 59

thank you!!!!!