Question

lim X-->infinity (e^4x +cosx -1) /2+e^x

Answer 1

I have you tried using squeeze theorem

Answer 2

we know -1<=cos(x)<=1
adding -1 on all sides gives
-2<=cos(x)-1<=0
adding e^(4x) on all sides gives
e^(4x)-2<=e^(4x)+cos(x)-1<=e^(4x)

then of course it will not change the direction of the inequality if you divide by a positive function e^x

Answer 3

so the answer is a range?

Answer 4

I didn't say that

Answer 5

I said to use squeeze theorem

Answer 6

that is why I set up an inequality

Answer 7

did you change your expression

Answer 8

earlier I seen
\[\lim_{x \rightarrow \infty}\frac{e^{4x}+\cos(x)-1}{e^{x}}\]

Answer 9

yes it was plus 2 at the bottome

Answer 10

i though it didnt matter

Answer 11

since after lhopital rule the 2 is gone

Answer 12

well e^x+2 is still positive
so divide all side of the inequality I have up there by e^x+2 instead of e^x

Answer 13

e^(4x)-2<=e^(4x)+cos(x)-1<=e^(4x) , all divided by (e^x +2)?

Answer 14

\[\frac{e^{4x}-2}{e^{x}+2} \le \frac{e^{4x}+\cos(x)-1}{e^{x}+2} \le \frac{e^{4x}}{e^x+2}\]
yep now apply squeeze theorem here

Answer 15

also called sandwich theorem or pinching theorem

Answer 16

infinity between infinity

Answer 17

\[\lim_{x \rightarrow \infty} \frac{e^{4x}}{e^x+2}=? \\ \lim_{x \rightarrow \infty}\frac{e^{4x}-2}{e^{4x}+2}=?\]

Answer 18

it is easier to find those limits

Answer 19

if those are the same then you determine the limit of the middle function

Answer 20

and yes that one is a type

Answer 21

\[\lim_{x \rightarrow \infty} \frac{e^{4x}}{e^x+2}=? \\ \lim_{x \rightarrow \infty}\frac{e^{4x}-2}{e^{x}+2}=? \]

Answer 22

i got infinity - infinity form for the limits

Answer 23

i see the cancellation of e^2x -2 , e^2x +2

Answer 24

since both have the same limit
then the middle function also goes to infinity as x goes to infinity

Answer 25

thanks

Answer 26

e^4x is much bigger than e^x so they both go to infinity?