Question

Another evaluate the limit

Answer 1

\[\sqrt{x^2+8x + 1}-x\]

Answer 2

After simplifying I got this \[\sqrt{8x + 1}\] am I on the right track or completely off?

Answer 3

as x tends to what ?

Answer 4

okay, I rationalize and get \[\frac{ x^2 + 8x +1 }{ \sqrt{x^2+8x+1} +x }\]

Answer 5

the limit is x approaches infinity

Answer 6

after simplifying? oh heck no

Answer 7

ratioanlizing should kill x^2 term on numerator

Answer 8

\[\sqrt{x^2+8x + 1}-x\times \frac{\sqrt{x^2+8x + 1}+x}{\sqrt{x^2+8x + 1}+x}\] is a start

Answer 9

oh i see you did that but made a slight mistake, as @ganeshie8 said

Answer 10

\[\frac{ 8x +1 }{ \sqrt{x^2+8x+1}+x }\]

Answer 11

\[\frac{ 8x +1 }{ \sqrt{x^2+8x+1} +x }\]

Answer 12

now eyeball it

Answer 13

looks good,
next you may factor x from the numerator and denominator

Answer 14

and kill it too

Answer 15

@satellite73 not your fan!! here is another example for you
Simplify the radical expression. 4×√(81x^20y^8

Answer 16

kill it all
this is just
\[\frac{8x}{2x}=4\]

Answer 17

Thanks guys!

Answer 18

there is the word "simplify" in what you need to do @triciaal
it is in the instruction "write in simplest radical form" which has a precise meaning

Answer 19

not found in most text books i have noticed, a serious oversight

Answer 20

If this was approaching -infinity would it still be 4?

Answer 21

not necessarily

Answer 22

do u get same value for e^x when x approaches both extreme values ?

Answer 23

That is a good question lol

Answer 24

your question is good too, so do you want to find the limit as x approaches -infty ?

Answer 25

yes

Answer 26

that would require no algebra, look at the original expression

Answer 27

try this :
substitute y = -x
as x->-infty,
y-> infty

Answer 28

Okay, I see. So it will be infinity.

Answer 29

\[\lim\limits_{x\to -\infty }\frac{ 8x +1 }{ \sqrt{x^2+8x+1}+x } = \lim\limits_{y\to +\infty }\frac{ 8(-y) +1 }{ \sqrt{(-y)^2+8(-y)+1}+(-y) } \]

Answer 30

does it really go to infinitty ?

Answer 31

even without the algebra, you get \(\infty +\infty\)

Answer 32

Yes it really does! lol

Answer 33

you need to show that it goes to infinity, eyeballing can be a dangerous game when working with limits

Answer 34

especially when u just start working with limits

Answer 35

but not in this case
\[\lim_{x\to -\infty}\sqrt{x^2+8x+1}=\infty \] and
\[\lim_{x\to -\infty}-x=\infty\]

Answer 36

it is not in an indeterminate form the the first one which would be \(\infty -\infty\) requiring more work

Answer 37

Okay, I believe I understand the concept, I can talk to my professor I little more about it tomorrow. Thanks for help @satellite73 and @ganeshie8