Question

Can you help me, please :
lim(-1)^n * x^n as x approches to +infinity

Answer 1

@amistre64 ?

Answer 2

sees loggy to me .... but thats just a thought.

what are we trying to determine?

Answer 3

\[\lim (-1)^{n} * x^{n} as x approaches \to + infinity\]

Answer 4

as x approaches infinity .... you sure thats not as n to infinity?

Answer 5

no. it's x

Answer 6

i dont see it converging for x >= 1

Answer 7

So ? @amistre64

Answer 8

im going to go out on a limb here and say ive got no idea what you are trying to ask for ...

Answer 9

@Michele_Laino

Answer 10

please wait a moment

Answer 11

If I call with a_n this quantity:
\[\Large {a_n} = {\left( { - 1} \right)^n}{x^n}\]

Answer 12

then we can write:
\[\Large {a_n} = {x^n}\]
if n is an even number
and its limits is +infinity
whereas, we have:
\[\Large {a_n} = - {x^n}\]
if n is an odd number, and its limit is -infinity
so what we can conclude?

Answer 13

I supposed that x>0, and x is different from zero

Answer 14

so the result would be.... ?

Answer 15

we conclude that the limit value doesn't exist

Answer 16

ok... what abou this :
lim\[\sin ^{2}x+\tan 2x + \sqrt{x}\] as x approaches to pi
Is sqrt of pi the result ?

Answer 17

we have:
sinx--->0, so (sinx)^2--->0*0=0
tan(2x)--->0
and:
sqrt(x)--->sqrt(pi)
so your answer is right!

Answer 18

what about
\[\lim (\sqrt{2x+1}-\sqrt{4x+1})/\sqrt[3]{x} as x approaches \to 0\]

Answer 19

we have to transform your function, as below:
\[\Large \begin{gathered}
\frac{{\sqrt {2x + 1} - \sqrt {4x + 1} }}{{\sqrt[3]{x}}} \times \frac{{\sqrt {2x + 1} + \sqrt {4x + 1} }}{{\sqrt {2x + 1} + \sqrt {4x + 1} }} = \hfill \\
\hfill \\
= \frac{{2x + 1 - 4x - 1}}{{\sqrt[3]{x}\left( {\sqrt {2x + 1} + \sqrt {4x + 1} } \right)}} = - \frac{{2\sqrt[3]{{{x^2}}}}}{{\sqrt {2x + 1} + \sqrt {4x + 1} }} \hfill \\
\end{gathered} \]

Answer 20

now, in order to compute your limit, you have to set x=0, into the last expression

Answer 21

Hmmm...right !
and the last one please
\[\lim (x^{n}+1)/x^{2} \] as x approaches to +infinity and n is in N

Answer 22

we can write your expression as follows:
\[\Large {x^{n - 2}} + \frac{1}{{{x^2}}}\]

Answer 23

now, if n<2, then n-2=-m<0, so your limit is equal to:
the limit of this function:
\[\frac{1}{{{x^m}}} + \frac{1}{{{x^2}}} \to ...?\]

Answer 24

please continue

Answer 25

what value do you get?

Answer 26

0?

Answer 27

that's right!

Answer 28

now, if n=2, then n-2=0, so your limit is the limit of this function:
\[\frac{1}{{{x^2}}} \to ...?\]

Answer 29

what do you get?

Answer 30

0 :D

Answer 31

that's right!

Answer 32

finally, if n>2, then n-2=q>0, so your limit is the limit of this function:
\[{x^q} + \frac{1}{{{x^2}}} \to ...?\]

Answer 33

what do you get?

Answer 34

what a minute ,, we have x^q ... that confuses me

Answer 35

yes! and q>0

Answer 36

infinity + 0 ?!

Answer 37

yes! and what is:
+infinity+0=...?

Answer 38

infinity ?

Answer 39

that's right!

Answer 40

so we have discovered that:
if n=2 or n<2, then your limit is zero,
whereas if n>2, then your limit is +infinity

Answer 41

If you could see my face right now, you'd see a look of gratitude :)

Answer 42

thanks! :)

Answer 43

I am indebted to you...Thank you so much :))))))

Answer 44

thanks again! :)

Answer 45

I'm very sorry, I have made an error:
in the case n=2, your limit is the limit of this function:
\[\Large {x^0} + \frac{1}{{{x^2}}} = 1 + \frac{1}{{{x^2}}} \to 1 + 0 = 1\]