Challenged in evaluating this limit.

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In summary, the conversation discusses an attempt to evaluate a limit using algebraic manipulation without using L'Hopital's rule. The original attempt using substitution did not lead to the desired result, but after further calculations, it was determined that the limit equals 0. The conversation also mentions a term in the limit that goes to 0 as n approaches infinity.
  • #1
futb0l
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I really haven't got a clue on how to evaluate this limit. I've tried doing algebraic manipulation, but to no avail. (L'Hopital's rule are not allowed to be used). If someone can give me a hint, that would be great :)

[tex]
\lim_{n\to\infty} \frac{(e^n)^2 + 1}{e^{n^2}+n}
[/tex]
 
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  • #2
I don't understand your notation on the denominator, is that e^(n^2) or what?
 
  • #3
futb0l said:
I really haven't got a clue on how to evaluate this limit. I've tried doing algebraic manipulation, but to no avail. (L'Hopital's rule are not allowed to be used). If someone can give me a hint, that would be great :)

[tex]
\lim_{n\to\infty} \frac{(e^n)^2 + 1}{e^{n^2}+n}
[/tex]

i am not quite sure on this, however i think it should go like this:

substitute e^n=t, than you get, ln e^n=lnt=>n=ln t, when n->infinity, t->infinity

lim(t^2+1)/(t^2+ln t),t->infinity

lim_t->infinity(t^2)/(t^2+lnt)+lim_t->infinity(1)/(t^2+lnt)
the second part is obviously 0 when t-> infinity, so we are left with

lim_t->infinity(t^2)/(t^2+lnt), we know that ln t=ln(1+(t-1)), which is equivalent with t-1, because the limit of their ratio is 1, we substitute it and we get

lim_t->infinity(t^2)/(t^2+t-1), so the limit of this is obviously 1.
 
  • #4
Oh ok I see your latex so I know what you mean. Give me a second to see if I can work it.
 
  • #5
Gib Z said:
I don't understand your notation on the denominator, is that e^(n^2) or what?

the denominator i enterpreted as (e^n)^2,
I do not know if this is what he meant?

Is this what you meant?
 
Last edited:
  • #6
I meant e^(n^2). So yeah, obviously the one you did with the 't' substitution is not what I'm looking for.
 
Last edited:
  • #7
than i guess the limit of that is going to be 0, since the denominator growes much faster than the numerator.
I will see if i can work it out now
 
  • #8
futb0l said:
I meant e^(n^2). So yeah, obviously the one you did with the 't' substitution is not what I'm looking for.

try it with the same substitution e^n=t, but just work it out in the form u meant. e^(ln^2 t), take this substitution then for e^(n^2), and i think you will get what u are looking for!
 
Last edited:
  • #9
sutupidmath said:
try it with the same substitution e^n=t, but just work it out in the form u meant. e^(ln t)^2, take this substitution then

Ok, so then you get:

[tex]
\lim_{t\to\infty} \frac{t^2 + 1}{e^{(\ln{t})^2} + \ln{t}}
[/tex]

So basically you will get infinity at the numerator and denominator, which doesn't lead to anything as far as I can see... *sigh*
 
Last edited:
  • #10
Denominator grows WAY faster than the numerator, so in this case it goes to 0
 
  • #11
Can you just say that without doing any calculations? o_O
 
  • #12
sutupidmath said:
than i guess the limit of that is going to be 0, since the denominator growes much faster than the numerator.
I will see if i can work it out now


i guess we cannot just say this, if this would be in an exam. however it is a clue to lead us on the right direction. i think we should do some calculations.

Although my approach i guess is obviously not appropriate(to much calculations).
 
  • #13
I think I just worked it out...

[tex]
\lim_{n\to\infty} \frac{ e^{2n} + 1 }{ e^{n^2} + n }
[/tex]

[tex]
\lim_{n\to\infty}\frac{ e^{2n}(1 + e^{-2n}) }{ e^{n^2}(1 + ne^{-n^2}) }
[/tex]

[tex]
\lim_{n\to\infty} e^{2n-n^2} \frac {(1 + e^{-2n}) }{(1 + ne^{-n^2}) }
[/tex]

[tex]
\lim_{n\to\infty} e^{2n-n^2} * 1
[/tex]

So that just equals to 0.
 
Last edited:
  • #14
futb0l said:
[tex]
\lim_{n\to\infty} e^{2n-n^2} \frac {(1 + e^{-2n}) }{(1 + ne^{-n^2}) }
[/tex]

[tex]
\lim_{n\to\infty} e^{2n-n^2} * 1
[/tex]

So that just equals to 0.

how do you know that the limit of the second part is 1
 
  • #15
Yeah, the ne^(-n^2) part is actually kind of part of the question. In the question it gives you limits of various functions - I didn't list it here because there is too many. So yeah, basically that term goes to 0 as n approaches infinite.
 

1. What is a limit in mathematics?

A limit in mathematics represents the value that a function or sequence approaches as its input or index approaches a certain value. It is used to describe the behavior of a function near a specific input value.

2. Why is evaluating a limit challenging?

Evaluating a limit can be challenging because it requires an understanding of the concept of infinity and how functions behave near certain input values. It also involves applying various mathematical techniques, such as algebraic manipulation or the use of calculus, to determine the value of the limit.

3. What are some common strategies for evaluating limits?

Some common strategies for evaluating limits include direct substitution, factoring, rationalization, and the use of limit laws. In more complex cases, techniques such as L'Hôpital's rule or Taylor series expansions may be used.

4. Can limits be evaluated at values where the function is undefined?

No, limits can only be evaluated at values where the function is defined. If the function is undefined at a certain input value, the limit at that value does not exist.

5. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of a function as the input value approaches from one side, either the left or the right. A two-sided limit, on the other hand, takes into account the behavior of the function from both the left and the right as the input value approaches a certain point.

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