$\displaystyle L=\lim_{x\to{0^{+}}}\bigg(a^{\frac{b}{x}+c}+f(x)+g (x)+...\bigg)^x$
Let $\displaystyle x = \frac{1}{t}$,
$\displaystyle \lim_{t\to {\infty}}
\bigg ( a^{bt+c} + f \bigg(\frac{1}{t}\bigg)+g\bigg(\frac{1}{t}\bigg)+. ..\bigg )^{\frac{1}{t}}$
$\displaystyle \lim_{t\to{\infty}}\bigg(a^{bt+c}\bigg)^{\frac{1}{ t}}\underbrace{\bigg(1+\frac{f\bigg(\frac{1}{t}\bi gg)}{a^{bt+c}}+\frac{g\bigg(\frac{1}{t}\bigg)}{a^{ bt+c}}\bigg)^{\frac{1}{t}}}_{1^0=1}$
Those fractions will vanish because $\displaystyle a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg( \frac{1}{x}\bigg)}\succ\cdots$
$\displaystyle \lim_{t\to{\infty}}\bigg(a^{bt} a^c\bigg)^{\frac{1}{t}}$
$\displaystyle \lim_{t\to{\infty}}a^b a^{\frac{c}{t}}$
$\displaystyle L = a^b$
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 $\displaystyle n\geq{0}$, and with positive leading
coefficients A and B. Evaluate
$\displaystyle L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt$
in terms of A and B and n
i'll solve a general case, i.e. for any positive real numbers $\displaystyle \alpha$ and $\displaystyle r < s: \ \lim_{x\to0+} \int_{rx}^{sx} \frac{\sin^{\alpha}t}{t^{\alpha + 1}} \ dt = \ln \left(\frac{s}{r}\right).$
choose $\displaystyle x$ small enough so that $\displaystyle 0 < rx < sx < \frac{\pi}{2}.$ then for any $\displaystyle rx < t < sx: \ 0< \sin t < t < \tan t.$
thus: $\displaystyle \frac{\cos^{\alpha}(sx)}{t} < \frac{\cos^{\alpha}t}{t}<\frac{\sin^{\alpha}t}{t^{ \alpha + 1}} < \frac{1}{t}.$ hence: $\displaystyle \cos^{\alpha}(sx)\int_{rx}^{sx} \frac{dt}{t} \leq \int_{rx}^{sx} \frac{\sin^{\alpha}t}{t^{\alpha + 1}} \ dt \leq \int_{rx}^{sx} \frac{dt}{t}.$ that is:
$\displaystyle \cos^{\alpha}(sx) \ln \left(\frac{s}{r} \right) \leq \int_{rx}^{sx} \frac{\sin^{\alpha} t}{t^{\alpha + 1}} \ dt \leq \ln \left(\frac{s}{r} \right).$ now since $\displaystyle \lim_{x\to 0+}\cos^{\alpha}(sx)=1,$ by the squeeze theorem:
$\displaystyle \lim_{x\to 0+} \int_{rx}^{sx} \frac{\sin^{\alpha} t}{t^{\alpha + 1}} \ dt = \ln \left(\frac{s}{r} \right). \ \ \ \square$
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 $\displaystyle 0\times \infty$. You can express the integral as a Riemann sum to see this indeterminacy.
it's very easy! the answer is $\displaystyle \frac{B}{(n+1)A}.$ so let $\displaystyle f(x)=Ax^{n+1} + ... , $ and $\displaystyle g(x)=Bx^n + ... .$
now we have: $\displaystyle L=\lim_{x\to\infty}g(x)e^{-f(x)} \int_0^x e^{f(t)} \ dt=\lim_{x\to\infty} \frac{\int_0^x e^{f(t)} \ dt}{\frac{e^{f(x)}}{g(x)}}.$ L'Hopital now gives us:
$\displaystyle L= \lim_{x\to\infty} \frac{(g(x))^2}{f'(x)g(x)-g'(x)}=\lim_{x\to\infty} \frac{B^2x^{2n} + \text{lower degree terms}}{(n+1)ABx^{2n} + \text{lower degree terms}}$
$\displaystyle =\frac{B}{(n+1)A}. \ \ \ \ \square$
$\displaystyle f(x)=ax^{n+1}+~...$
$\displaystyle g(x)=bx^{n}+~...$
$\displaystyle L=\lim_{x\to \infty}g(x)\int_0^x e^{f(t)-f(x)}dt$
$\displaystyle L=\lim_{x\to \infty}\frac{g(x)\int_0^x e^{f(t)}dt}{e^{f(x)}}$
Use L'hopital, (I'm not in the mood to do it the pro way, sorry)
$\displaystyle L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]+g(x)\cdot e^{f(x)}}{e^{f(x)}f'(x)}$
$\displaystyle L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{g(x)\cdot e^{f(x)}}{e^{f(x)}f'(x)}$
$\displaystyle L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{g(x)}{f'(x)}$
$\displaystyle L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{b}{a(n+1)}$
Now it can easily be shown that the other limit goes to zero. Simplify $\displaystyle \frac{g'(x)}{f'(x)}$ and use L'hopital. Then only the other fraction is left.
So,
$\displaystyle L=\frac{b}{a(n+1)}$