Question

find the limit

Answer 1

\[\lim_{x \rightarrow \infty} (\sqrt{16x^2+x}-4x)\]

Answer 2

there is no denominator?
if no - lim= infinity (no limit)

Answer 3

ok I am going to try to remember this...

so if no denominator then equals infinity.

RIGHT?

Answer 4

it say it is wrong

Answer 5

just think about it...
x->infinity...
only numbers you have -> infinity
you can't cancel out anything...
has to be infinity, right?

Answer 6

wrong?
let's check the original problem...
is it: sq rt (16x^2+x)- 4x ?
give me a min

Answer 7

oup! I see

Answer 8

I have four more i need help with....I am going to switch to another computer..Please can you help me with these last 4 problems...

Answer 9

let's do the following:
\[\lim(\sqrt{16x^{2}+x}-4x)*(\sqrt{16x ^{2}+x}+4x)/(\sqrt{16x ^{2}+x}+4x)\]
do you see now?

Answer 10

ok i will be right back

Answer 11

\[=\lim(16x ^{2}+4-16x ^{2})/(\sqrt{16x ^{2}+x}+4x)<=\lim x/(4x+4x)=1/8\]
=1/8
are we good now?

Answer 12

i'll be back in a few too :)

Answer 13

this one:
\[\lim(x+\sqrt{x^2+5x})\]
use the same approach:
multiply & divide by
\[x-\sqrt{x^2+5x}\]

Answer 14

you'll get
\[\lim(-5x)/(x-\sqrt{x^2+5x})=\lim 5/(-1+\sqrt{1+5/x}=5/(\lim(-1)+\lim(\sqrt{1+5/x})\]

Answer 15

\[=5/(\lim(-1)+\lim \sqrt{(1+5/x)})=...-5/2\]

Answer 16

this is another problem (3) from your list:
lim(x^2+x^3)=
lim x^2(1+x)=limx^2 *(lim1 + limx)=-limx^2 * infinity=-infinity

Answer 17

lim arctan (x^4-x^9)=-lim arctan(x^9-x^4)=-pi/2