This Limit problem seems too simple....

[PeroK]

further on this point, for any $x \geq 0$
$\sqrt{x^6+1}= \sqrt{(x^3)^2+1^2}\leq \sqrt{(x^3)^2}+\sqrt{1^2} = x^3 + 1$
this is triangle inequality... negating then gives the 'desired direction'

learning to apply triangle inequality seems far more important than blasting this with l'hopital, though I suppose I don't know what is actually emphasized in calc 101

You don't even need the triangle inequality. Simply:

$(x^3 + 1)^2 = x^6 + 2x^3 + 1 > x^6 + 1 \ \ (x > 0)$

$x^3 + 1 > \sqrt{x^6+1}$

 

[Wi_N]

We only need to do it once then simplify the $x$s in this case (remember that $a^na^m=a^{n+m}$ and that $\frac{1}{a^n}=a^{-n}$).
$$\lim_{x\to\infty}\frac{-x^4}{-3x^2}e^\frac{1}{x}=\lim_{x\to\infty}\frac{x^2}{3}e^\frac{1}{x}=\infty$$

is there a specific reason u kept the higher terms and removed the lower terms? is there a reasoning?

 

[PeroK]

is there a specific reason u kept the higher terms and removed the lower terms? is there a reasoning?

Let $f(x) = g(x) - h(x)$

Suppose we can show that $g(x) - h(x)$ diverges to $+\infty$, then we know that $f(x)$ diverges.

Now, suppose we find $g_2(x) < g(x)$, then:

$f(x) > g_2(x) - h(x)$

And, if we can see that $g_2(x) - h(x)$ diverges to $+\infty$, then we know that $f(x)$ diverges.

Finally, if we can find $h_2(x) > h(x)$, then:

$f(x) > g_2(x) - h_2(x)$

And, if we can see that $g_2(x) - h_2(x)$ diverges to $+\infty$, then we know that $f(x)$ diverges.

That's the idea.

 

[archaic]

is there a specific reason u kept the higher terms and removed the lower terms? is there a reasoning?

I do not think that this is rigorous but it is intuitive.
$$
\lim_{x\to\infty}\frac{2x^3+x+6}{x^3}
=\lim_{x\to\infty}\frac{x^3(2+\underbrace{\frac{x}{x^3}}_\text{goes to 0}+\underbrace{\frac{6}{x^3}}_\text{goes to 0})}{x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}\frac{2x^3+x+6}{x^3}=2\\
\Leftrightarrow \frac{\lim_{x\to\infty}2x^3+x+6}{\lim_{x\to\infty}x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=2\lim_{x\to\infty}x^3\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=\lim_{x\to\infty}2x^3
$$

 

[PeroK]

I do not think that this is rigorous but it is intuitive.
$$
\lim_{x\to\infty}\frac{2x^3+x+6}{x^3}
=\lim_{x\to\infty}\frac{x^3(2+\underbrace{\frac{x}{x^3}}_\text{goes to 0}+\underbrace{\frac{6}{x^3}}_\text{goes to 0})}{x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}\frac{2x^3+x+6}{x^3}=2\\
\Leftrightarrow \frac{\lim_{x\to\infty}2x^3+x+6}{\lim_{x\to\infty}x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=2\lim_{x\to\infty}x^3\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=\lim_{x\to\infty}2x^3
$$

In general, this doesn't hold. Especially:
$$
\lim_{x\to\infty}\frac{2x^3+x+6}{x^3}=2\\
\Rightarrow \frac{\lim_{x\to\infty}2x^3+x+6}{\lim_{x\to\infty}x^3}=2\\
$$
is false.

 

[Wi_N]

I do not think that this is rigorous but it is intuitive.
$$
\lim_{x\to\infty}\frac{2x^3+x+6}{x^3}
=\lim_{x\to\infty}\frac{x^3(2+\underbrace{\frac{x}{x^3}}_\text{goes to 0}+\underbrace{\frac{6}{x^3}}_\text{goes to 0})}{x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}\frac{2x^3+x+6}{x^3}=2\\
\Leftrightarrow \frac{\lim_{x\to\infty}2x^3+x+6}{\lim_{x\to\infty}x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=2\lim_{x\to\infty}x^3\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=\lim_{x\to\infty}2x^3
$$

problem i had was that they told me i can't eliminate terms unless I am transitioning fully off the limit. mixed signals.

 

[PeroK]

problem i had was that they told me i can't eliminate terms unless I am transitioning fully off the limit. mixed signals.

You're right. That post was not correct.

 

[archaic]

You're right. That post was not correct.

My bad, it is because the limit doesn't exist that we can't separate the limit, right?

 

[PeroK]

My bad, it is because the limit doesn't exist that we can't separate the limit, right?

Limits may exists as real numbers and limits may exist as $\pm\infty$. And, limits may be of an indefinite form, including $0/0$ and $\infty/\infty$.

You need to take care.

 

[archaic]

Limits may exists as real numbers and limits may exist as $\pm\infty$. And, limits may be of an indefinite form, including $0/0$ and $\infty/\infty$.

You need to take care.

Why was that manipulation wrong though?
http://tutorial.math.lamar.edu/Classes/CalcI/LimitsProperties.aspx

 

[PeroK]

Why was that manipulation wrong though?

Let's replace the specific functions with some general ones to see what you are saying:

$$
\lim_{x\to\infty}\frac{f(x)}{g(x)}=2\\
(1) \Leftrightarrow \frac{\lim_{x\to\infty}f(x)}{\lim_{x\to\infty}g(x)}=2\\
(2) \Leftrightarrow \lim_{x\to\infty}f(x)=2\lim_{x\to\infty}g(x)\\
(3) \Leftrightarrow \lim_{x\to\infty}f(x)=\lim_{x\to\infty}2g(x)
$$

If we call these steps (1), (2) and (3):

Step (1) is not true, because the limits might not exist. E.g. alternating functions. Plus, you now may have an indeterminate form that is no longer under a limit process. In the specific case you have:
$$\frac{+\infty}{+\infty} = 2$$
Which is not a valid statement.

Step (2) does not hold if the limits are infinite or $0$.

Step (3) does not hold if the limits are infinite.

 

[archaic]

Let's replace the specific functions with some general ones to see what you are saying:
If we call these steps (1), (2) and (3):

Step (1) is not true, because the limits might not exist. E.g. alternating functions. Plus, you now may have an indeterminate form that is no longer under a limit process. In the specific case you have:
$$\frac{+\infty}{+\infty} = 2$$
Which is not a valid statement.

Step (2) does not hold if the limits are infinite or $0$.

Step (3) does not hold if the limits are infinite.

Thank you! So do we know that the limit of a polynomial is that of its term of highest degree only because $\lim_{x\to\infty}c_nx^n+c_{n-1}x^{n-1}+...+c_0=\lim_{x\to\infty}x^n(c_n+\underbrace{\frac{c_{n-1}}{x}+...+\frac{c_0}{x^n}}_\text{=0})$ and without further steps?

 

[PeroK]

Thank you! So do we know that the limit of a polynomial is that of its term of highest degree only because $\lim_{x\to\infty}c_nx^n+c_{n-1}x^{n-1}+...+c_0=\lim_{x\to\infty}x^n(c_n+\underbrace{\frac{c_{n-1}}{x}+...+\frac{c_0}{x^n}}_\text{=0})$ and without further steps?

Why you really want is:

$\lim_{x\to\infty}\frac{c_nx^n+c_{n-1}x^{n-1}+...+c_0}{x^n} = c_n$

 

[PeroK]

Thank you! So do we know that the limit of a polynomial is that of its term of highest degree only because $\lim_{x\to\infty}c_nx^n+c_{n-1}x^{n-1}+...+c_0=\lim_{x\to\infty}x^n(c_n+\underbrace{\frac{c_{n-1}}{x}+...+\frac{c_0}{x^n}}_\text{=0})$ and without further steps?

This is fairly meaningless, as $c_n x^n$ is not an asymptote for your polynomial. I.e.:

$\lim_{x\to\infty}c_nx^n+c_{n-1}x^{n-1}+...+c_0=\lim_{x\to\infty}x^n(c_n+\underbrace{\frac{c_{n-1}}{x}+...+\frac{c_0}{x^n}}_\text{=0}) = \lim_{x\to\infty}c^nx^n = \lim_{x\to\infty}x = +\infty$

Doesn't tell you a lot. For example:

$\lim_{x\to\infty}((c_nx^n+c_{n-1}x^{n-1}+...+c_0) - c_nx^n) \ne 0$

 

[archaic]

This is fairly meaningless, as $c_n x^n$ is not an asymptote for your polynomial. I.e.:

$\lim_{x\to\infty}c_nx^n+c_{n-1}x^{n-1}+...+c_0=\lim_{x\to\infty}x^n(c_n+\underbrace{\frac{c_{n-1}}{x}+...+\frac{c_0}{x^n}}_\text{=0}) = \lim_{x\to\infty}c^nx^n = \lim_{x\to\infty}x = +\infty$

Doesn't tell you a lot. For example:

$\lim_{x\to\infty}((c_nx^n+c_{n-1}x^{n-1}+...+c_0) - c_nx^n) \ne 0$

But why is it that we can substitute a polynomial by its $c_nx^n$ in some limits? In your example you would just cancel both $n$th degree terms and then do the same manipulation with $x^{n-1}$.

 

[PeroK]

But why is it that we can substitute a polynomial by its $c_nx^n$ in some limits? In your example you would just cancel both $n$th degree terms and then do the same manipulation with $x^{n-1}$.

The meaningful statement is:

$\lim_{x\to\infty}\frac{c_nx^n+c_{n-1}x^{n-1}+...+c_0}{x^n} = c_n$

That's what allows you to do stuff.

 

[archaic]

The meaningful statement is:
That's what allows you to do stuff.

But how do you go from that to $\lim_{x\to\infty}P(x)=\lim_{x\to\infty}c_nx^n$ inside another limit?
Do you $\lim_{x\to\infty}P(x)G(x)=\lim_{x\to\infty}\frac{P(x)}{x^n}x^nG(x)=\lim_{x\to\infty}c_nx^nG(x)$?

 

[PeroK]

But how do you go from that to $lim_{x\to\infty}P(x)=lim_{x\to\infty}c_nx^n$?
Do you $lim_{x\to\infty}P(x)=lim_{x\to\infty}\frac{P(x)}{x^n}x^n=lim_{x\to\infty}c_nx^n$?

The first statement is really just saying both limits are +infinity. If you are working loosely, then you can throw stuff around. But, that's not what we are doing here.

The problem you have is that $P(x)$ and $c_n x^n$ do not converge. You need to avoid things like:

$lim_{x\to\infty}P(x)=lim_{x\to\infty}c_nx^n$

Then:

$lim_{x\to\infty} (P(x) - c_nx^n) = 0$

This is beside the point. You can't just throw things around, not worrying whether your limits are finite, infinite or 0.

 

[archaic]

The first statement is really just saying both limits are +infinity. If you are working loosely, then you can throw stuff around. But, that's not what we are doing here.

The problem you have is that $P(x)$ and $c_n x^n$ do not converge. You need to avoid things like:

$lim_{x\to\infty}P(x)=lim_{x\to\infty}c_nx^n$

Then:

$lim_{x\to\infty} (P(x) - c_nx^n) = 0$

This is beside the point. You can't just throw things around, not worrying whether your limits are finite, infinite or 0.

Oh thank you, but I update my post while you were answering, I think.

 

[vela]

But why is it that we can substitute a polynomial by its $c_nx^n$ in some limits?

Generally, you can't because cancellation may occur. If you can see that cancellation won't occur, then you can do the replacement. In this thread's problem, the leading order term is going to cancel, so it's not obvious you can simply discard the lower-order terms in the polynomial.

 

[vela]

The Calc 1 approach to this problem is probably to multiply and divide by the conjugate:
\begin{align*}
\left(x^3 +x^2 +\frac{x}{2}\right)e^{\frac{1}{x}} - \sqrt{x^6+1}
&= \left[\left(x^3 +x^2 +\frac{x}{2}\right)e^{\frac{1}{x}} - \sqrt{x^6+1}\right]\frac{\left(x^3 +x^2 +\frac{x}{2}\right)e^{\frac{1}{x}} + \sqrt{x^6+1}}{\left(x^3 +x^2 +\frac{x}{2}\right)e^{\frac{1}{x}} + \sqrt{x^6+1}} \\
&= \frac{\left(x^3 +x^2 +\frac{x}{2}\right)^2e^{\frac{2}{x}} - (x^6+1)}{\left(x^3 +x^2 +\frac{x}{2}\right)e^{\frac{1}{x}} + \sqrt{x^6+1}} \\
\end{align*} It's fairly straightforward to analyze the problem from here with the usual Calc 1 methods.

 

[archaic]

Generally, you can't because cancellation may occur. If you can see that cancellation won't occur, then you can do the replacement. In this thread's problem, the leading order term is going to cancel, so it's not obvious you can simply discard the lower-order terms in the polynomial.

So how can generally foresee that it is not OK to substitute a function with its asymptotic equivalent?

 

[fresh_42]

So how can generally foresee that it is not OK to substitute a function with its asymptotic equivalent?

It is ok. You simply have to consider all factors, quotients, sums and differences to determine what asymptotic really is!