# Math Help - Limit Marathon

1. Originally Posted by Mathstud28
Here is another one I came up with, find L.

$L=\lim_{x\to{0^{+}}}\bigg(a^{\frac{b}{x}+c}+f(x)+g (x)+...\bigg)^x$

Where $a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg( \frac{1}{x}\bigg)}\succ\cdots$

Restrictions: You may not use asymptotic equivalences, rate of increase, or in any other way disregard terms.

EDIT: typo, also assuming there is no domain conflictions with other functions
Hint: Let $x=\frac{1}{\varphi}$

2. Originally Posted by Mathstud28
Here is another one I came up with, find L.

$L=\lim_{x\to{0^{+}}}\bigg(a^{\frac{b}{x}+c}+f(x)+g (x)+...\bigg)^x$

Where $a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg( \frac{1}{x}\bigg)}\succ\cdots$

Restrictions: You may not use asymptotic equivalences, rate of increase, or in any other way disregard terms.

EDIT: typo, also assuming there is no domain conflictions with other functions
$L=\lim_{x\to{0^{+}}}\bigg(a^{\frac{b}{x}+c}+f(x)+g (x)+...\bigg)^x$

Let $x = \frac{1}{t}$,

$\lim_{t\to {\infty}}
\bigg ( a^{bt+c} + f \bigg(\frac{1}{t}\bigg)+g\bigg(\frac{1}{t}\bigg)+. ..\bigg )^{\frac{1}{t}}$

$\lim_{t\to{\infty}}\bigg(a^{bt+c}\bigg)^{\frac{1}{ t}}\underbrace{\bigg(1+\frac{f\bigg(\frac{1}{t}\bi gg)}{a^{bt+c}}+\frac{g\bigg(\frac{1}{t}\bigg)}{a^{ bt+c}}\bigg)^{\frac{1}{t}}}_{1^0=1}$

Those fractions will vanish because $a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg( \frac{1}{x}\bigg)}\succ\cdots$

$\lim_{t\to{\infty}}\bigg(a^{bt} a^c\bigg)^{\frac{1}{t}}$

$\lim_{t\to{\infty}}a^b a^{\frac{c}{t}}$

$L = a^b$

3. Originally Posted by wingless
$L=\lim_{x\to{0^{+}}}\bigg(a^{\frac{b}{x}+c}+f(x)+g (x)+...\bigg)^x$

Let $x = \frac{1}{t}$,

$\lim_{t\to {\infty}}
\bigg ( a^{bt+c} + f \bigg(\frac{1}{t}\bigg)+g\bigg(\frac{1}{t}\bigg)+. ..\bigg )^{\frac{1}{t}}$

$\lim_{t\to{\infty}}\bigg(a^{bt+c}\bigg)^{\frac{1}{ t}}\underbrace{\bigg(1+\frac{f\bigg(\frac{1}{t}\bi gg)}{a^{bt+c}}+\frac{g\bigg(\frac{1}{t}\bigg)}{a^{ bt+c}}\bigg)^{\frac{1}{t}}}_{1^0=1}$

f and g will vanish because $a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg( \frac{1}{x}\bigg)}\succ\cdots$

$\lim_{t\to{\infty}}\bigg(a^{bt} a^c\bigg)^{\frac{1}{t}}$

$\lim_{t\to{\infty}}a^b a^{\frac{c}{t}}$

$L = a^b$
Almost identical to my solution ...now come on Wingless let's see your limit problem.

4. Mathstud, Are we restricted to post limits that have only one variable? Could we post limits for multivariable functions?

5. Originally Posted by Chris L T521
Mathstud, Are we restricted to post limits that have only one variable? Could we post limits for multivariable functions?
I really did not intend it to be so, for if you cannot find limits in one variable what hope do you have in two? (You who hasn't answered any single variable limits yet )

Don't you have any single variable ones?

6. Originally Posted by Mathstud28
Don't you have any single variable ones?
The ones I have are too easy... . I'll think of a couple hard ones...

7. Ok then, here is an easy one:

$\lim_{x\to 0^+} \int_x^{2x}\frac{\sin^m t}{t^{m+1}}~dt,~m \in \mathbb{N}^+$

8. Originally Posted by wingless
Ok then, here is an easy one:

$\lim_{x\to 0^+} \int_x^{2x}\frac{\sin^m t}{t^{m+1}}~dt,~m \in \mathbb{N}^+$
Hmm...Maybe I am crazy or stupid or a combintation of both, but

$\lim_
{x\to{0^+}}\int_x^{2x}\frac{\sin^m(t)}{t^{m+1}}dt= \int_0^0\frac{\sin^m(t)}{t^{m+1}}dt=0$

Is there a hidden indetermincay I am missing?

9. Here's one I puzzled with a while. You all will probably wade through it like Grant through Richmond.

No restrictions. Use L'Hopital, whatever. I sure don't care.

Let f(x) and g(x) be polynomials with real coefficients of degree n+1, n

respectively, where $n\geq{0}$, and with positive leading

coefficients A and B. Evaluate

$L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt$

in terms of A and B and n

10. Originally Posted by galactus
Here's one I puzzled with a while. You all will probably wade through it like Grant through Richmond.

No restrictions. Use L'Hopital, whatever. I sure don't care.

Let f(x) and g(x) be polynomials with real coefficients of degree n+1, n

respectively, where $n\geq{0}$, and with positive leading

coefficients A and B. Evaluate

$L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt$

in terms of A and B and n
so is it

$f(x)=ax^{n+1}+bx^{n}+\cdots$

or is it

$f(x)=ax^{n+1}+bx^n$?

11. Originally Posted by wingless
Ok then, here is an easy one:

$\lim_{x\to 0^+} \int_x^{2x}\frac{\sin^m t}{t^{m+1}}~dt,~m \in \mathbb{N}^+$
i'll solve a general case, i.e. for any positive real numbers $\alpha$ and $r < s: \ \lim_{x\to0+} \int_{rx}^{sx} \frac{\sin^{\alpha}t}{t^{\alpha + 1}} \ dt = \ln \left(\frac{s}{r}\right).$

choose $x$ small enough so that $0 < rx < sx < \frac{\pi}{2}.$ then for any $rx < t < sx: \ 0< \sin t < t < \tan t.$

thus: $\frac{\cos^{\alpha}(sx)}{t} < \frac{\cos^{\alpha}t}{t}<\frac{\sin^{\alpha}t}{t^{ \alpha + 1}} < \frac{1}{t}.$ hence: $\cos^{\alpha}(sx)\int_{rx}^{sx} \frac{dt}{t} \leq \int_{rx}^{sx} \frac{\sin^{\alpha}t}{t^{\alpha + 1}} \ dt \leq \int_{rx}^{sx} \frac{dt}{t}.$ that is:

$\cos^{\alpha}(sx) \ln \left(\frac{s}{r} \right) \leq \int_{rx}^{sx} \frac{\sin^{\alpha} t}{t^{\alpha + 1}} \ dt \leq \ln \left(\frac{s}{r} \right).$ now since $\lim_{x\to 0+}\cos^{\alpha}(sx)=1,$ by the squeeze theorem:

$\lim_{x\to 0+} \int_{rx}^{sx} \frac{\sin^{\alpha} t}{t^{\alpha + 1}} \ dt = \ln \left(\frac{s}{r} \right). \ \ \ \square$

12. Originally Posted by Mathstud28
Hmm...Maybe I am crazy or stupid or a combintation of both, but

$\lim_
{x\to{0^+}}\int_x^{2x}\frac{\sin^m(t)}{t^{m+1}}dt= \int_0^0\frac{\sin^m(t)}{t^{m+1}}dt=0$

Is there a hidden indetermincay I am missing?
Yes. Think of the Riemann sum, and try to see the area under the curve as a rectangle. When x approaches 0, the rectangle will have no width and infinite height. So its area is $0\times \infty$. You can express the integral as a Riemann sum to see this indeterminacy.

13. Originally Posted by galactus
Here's one I puzzled with a while. You all will probably wade through it like Grant through Richmond.

No restrictions. Use L'Hopital, whatever. I sure don't care.

Let f(x) and g(x) be polynomials with real coefficients of degree n+1, n

respectively, where $n\geq{0}$, and with positive leading

coefficients A and B. Evaluate

$L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt$

in terms of A and B and n
it's very easy! the answer is $\frac{B}{(n+1)A}.$ so let $f(x)=Ax^{n+1} + ... ,$ and $g(x)=Bx^n + ... .$

now we have: $L=\lim_{x\to\infty}g(x)e^{-f(x)} \int_0^x e^{f(t)} \ dt=\lim_{x\to\infty} \frac{\int_0^x e^{f(t)} \ dt}{\frac{e^{f(x)}}{g(x)}}.$ L'Hopital now gives us:

$L= \lim_{x\to\infty} \frac{(g(x))^2}{f'(x)g(x)-g'(x)}=\lim_{x\to\infty} \frac{B^2x^{2n} + \text{lower degree terms}}{(n+1)ABx^{2n} + \text{lower degree terms}}$

$=\frac{B}{(n+1)A}. \ \ \ \ \square$

14. Originally Posted by galactus
Here's one I puzzled with a while. You all will probably wade through it like Grant through Richmond.

No restrictions. Use L'Hopital, whatever. I sure don't care.

Let f(x) and g(x) be polynomials with real coefficients of degree n+1, n

respectively, where $n\geq{0}$, and with positive leading

coefficients A and B. Evaluate

$L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt$

in terms of A and B and n

$f(x)=ax^{n+1}+~...$

$g(x)=bx^{n}+~...$

$L=\lim_{x\to \infty}g(x)\int_0^x e^{f(t)-f(x)}dt$

$L=\lim_{x\to \infty}\frac{g(x)\int_0^x e^{f(t)}dt}{e^{f(x)}}$

Use L'hopital, (I'm not in the mood to do it the pro way, sorry)

$L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]+g(x)\cdot e^{f(x)}}{e^{f(x)}f'(x)}$

$L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{g(x)\cdot e^{f(x)}}{e^{f(x)}f'(x)}$

$L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{g(x)}{f'(x)}$

$L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{b}{a(n+1)}$

Now it can easily be shown that the other limit goes to zero. Simplify $\frac{g'(x)}{f'(x)}$ and use L'hopital. Then only the other fraction is left.

So,
$L=\frac{b}{a(n+1)}$

15. Of course, that is it. I looked at it for a little while before it dawned on me to use L'Hopital, then it falls into place pretty easy.

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