hi,I need the answers of quuestions given below (one of my colleague asks, I could not find at its first look line)

 

1) limx→∞(2−x+1)x

2) the graph for 2xx2+1

 

sorry I could not set the formula in the new version of this website. 1) lim x to infinity (2^-x+1)^x

2) the graph for 2x / (x^2+1)

Edited June 9, 20223 yr by ahmet

Theres a great graphing tool online called "Desmos" which would allow for you to make such graphs. As for ze limit there are also other online resources you can use to solve it. We can't do ze work for ya. Only provide examples tools and give assistance in conceptual issues and check process ta help ya learn.

Aline I did not understand what you said well , what does ze and ya mean?

 

note please I am mathematics teacher and the asker of this question is another maths teacher.

 

this is not a homework help request. 

 

in the second question ,the asker is probably asking the graph with quick analysis (i.e. local minimum/maximum points , asymptotes etc) or I suppose so.

thanks

Edited June 9, 20223 yr by ahmet

This is what you want:

https://en.wikipedia.org/wiki/Logarithmic_differentiation#:~:text=In calculus%2C logarithmic differentiation or,rather than the function itself.

Plus changing exponential bases: 2^(-x)=e^(-xln2)

7 minutes ago, joigus said:

This is what you want:

https://en.wikipedia.org/wiki/Logarithmic_differentiation#:~:text=In calculus%2C logarithmic differentiation or,rather than the function itself.

Plus changing exponential bases: 2^(-x)=e^(-xln2)

Hi, what is your result if this works

Sorry. I did the derivative. You need the limit. The limit is of the kind 1^(infinity), which are solved by relating them to the definition of e=lim(1+f)^(1/f) if f-> 0 when x->infinity.

 

The result is 1, if I'm not mistaken.

I'll take another look tomorrow. Tell me what you get.

3 hours ago, joigus said:

Sorry. I did the derivative. You need the limit. The limit is of the kind 1^(infinity), which are solved by relating them to the definition of e=lim(1+f)^(1/f) if f-> 0 when x->infinity.

 

The result is 1, if I'm not mistaken.

I'll take another look tomorrow. Tell me what you get.

I tried this,  unfortunately i could not reach a meaningful result because of the persistence of ambiguity

Edited June 10, 20223 yr by ahmet

Hi once again the second question resolved, the first one is also resolved but i could not be sure for that one. 

Hi. I did not address the second question, because I figured you would solve it as soon as you applied yourself to it.

Apparently you've solved the 1st one but are unsure of its validity?

The first question is best addressed with assymptotics.

Here's a sketch:

\[\left(1+2^{-x}\right)^{x}\sim\left(1+2^{-x}\right)^{2^{x}x2^{-x}}\]

Now,

\[\left(1+2^{-x}\right)^{2^{x}}\sim e\]

And,

\[x2^{-x}\sim0\]

So,

\[\left(1+2^{-x}\right)^{x}\sim e^{0}=1\]

Now, making it rigorous is another matter. In fact, I went dangerously off-bounds when I said \(x2^{-x}\sim0\), as you're not supposed to ever say anything is ∼0 in assymptotics.

Edited June 10, 20223 yr by joigus
correcting for bad LateX rendering

2 hours ago, joigus said:

Hi. I did not address the second question, because I figured you would solve it as soon as you applied yourself to it.

Apparently you've solved the 1st one but are unsure of its validity?

The first question is best addressed with assymptotics.

Here's a sketch:

 

(1+2−x)x∼(1+2−x)2xx2−x

 

Now,

 

(1+2−x)2x∼e

 

And,

 

x2−x∼0

 

So,

 

(1+2−x)x∼e0=1

 

Now, making it rigorous is another matter. In fact, I went dangerously off-bounds when I said x2−x∼0 , as you're not supposed to ever say anything is ∼0 in assymptotics.

really , I tried to benefit from (1+1/n)^n=e (n--->infinity)

but reached to the same result and as you mention I could not be sufficiently sure for the limit appearing in the last line.

Is what you want anywhere at the level of rigorousness as proving that,

\[ \forall\varepsilon>0\:\exists N>0\:/\:x>N\Rightarrow\left|\left(1+2^{-x}\right)^{x}-1\right|<\varepsilon \]

?

28 minutes ago, joigus said:

Is what you want anywhere at the level of rigorousness as proving that,

 

∀ε>0∃N>0/x>N⇒∣∣(1+2−x)x−1∣∣<ε

Ah yes this was good! but...

shall we be able to find the delta correlated to epsilon here?

Edited June 10, 20223 yr by ahmet

Wouldn't you be satisfied with the regular lemmas and theorems, like L'Hôpital and such? Finding an N(epsilon) is a pain in the neck.  

27 minutes ago, joigus said:

Wouldn't you be satisfied with the regular lemmas and theorems, like L'Hôpital and such? Finding an N(epsilon) is a pain in the neck.  

hahahaha haha ha ha