I am completly blocked trying to prove the solution of the limits below
1) $$ \lim _{m\to\infty}\left(\cos\left(\frac{x}{m}\right)\right)^m\\1 \quad \text{ for } a\to +\infty;\quad 0 \quad \text{ for } \quad a \to -\infty $$
2)$$ \lim _{a\to\infty}\frac{a^x-1}{x}, a>1\\+\infty\, \text{ for }\, x\to +\infty;0\, \text{ for }\, a \to -\infty $$
I believe both limits are related to $\lim_{x\to\infty}(1+\frac{1}{x})^x =e$, but I just can't find a way to get there.
In case 1) I get $(1^+)^\infty$ which I can't simplify and get to a clear limit
In case 2) using the change of variable $y = a^x-1$, I can get to
$$ \lim _{a\to\infty}\frac{a^x-1}{x} = \lim _{a\to\infty}\frac{\ln(a)}{\ln(1+y)^\frac{1}{y}} $$
but I then obtain a $\infty^\frac{1}{\infty}$, which, again, I can't solve.
Can someone give some hints on what I am missing here? Or am I using a completly wrong approach?
$\endgroup$ 44 Answers
$\begingroup$For (1), take a logarithm to get$$ \lim_{m \to \infty} \frac{\log \cos\left( \frac{x}{m} \right)}{\frac{1}{m}}, $$which by L'Hôpital is$$ \lim_{m \to \infty}\frac{-\sin \left( \frac{x}{m} \right)\frac{x}{m^2}}{\cos\left(\frac{x}{m}\right)\frac{1}{m^2}}, $$which simplifies to$$ \lim_{m \to \infty} -x\tan\left(\frac{x}{m}\right), $$which is $0$, so the answer is $1$ since we took a logarithm.
For (2), the limit of the given function is clearly infinite, but I suspect there should be an $a$ in the denominator instead of an $x$. Then you can use L'Hôpital directly to get$$ \lim_{a \to \infty} xa^{x-1} $$which is $1$ if $x = 1$, $0$ if $x < 1$, infinite otherwise.
$\endgroup$ 1 $\begingroup$I'm taught to do $1^\infty$ only such way:
$$\begin{align*}
\lim\limits_{m\to\infty}\left(\cos\left(\frac{x}{m}\right)\right)^m &=
\lim\limits_{m\to\infty}\left(1+\cos\left(\frac{x}{m}\right)-1\right)^m\\
&=\lim\limits_{m\to\infty}\left(1+\cos\left(\frac{x}{m}\right)-1\right)^
{\frac{1}{\cos\left(\frac{x}{m}\right)-1}\cdot
\left(\cos\left(\frac{x}{m}\right)-1\right)m}\\
&=\left(
\lim\limits_{m\to\infty}\left(1+\cos\left(\frac{x}{m}\right)-1\right)^
\frac{1}{\cos\left(\frac{x}{m}\right)-1}\right)^{
\lim\limits_{m\to\infty}
\left(\cos\left(\frac{x}{m}\right)-1\right)m}\tag{1}\\
&=e^{
\lim\limits_{m\to\infty}
\left(\cos\left(\frac{x}{m}\right)-1\right)m};
\end{align*}$$$$\begin{align*}
\lim\limits_{m\to\infty}
\left(\cos\left(\frac{x}{m}\right)-1\right)m
&=
\lim\limits_{m\to\infty}
\left(1-2\sin^2\left(\frac{x}{2m}\right)-1\right)m\\
&=-2\lim\limits_{m\to\infty}
\sin^2\left(\frac{x}{2m}\right)\cdot\frac{4m^2}{x^2}\cdot\frac{x^2}{4m}\\
&=-\frac{x^2}{2}\left(\lim\limits_{m\to\infty}
\sin^2\left(\frac{x}{2m}\right)\cdot\frac{4m^2}{x^2}\right)\cdot
\left(\lim\limits_{m\to\infty}\frac{1}{m}\right)\tag{2}\\
&=-\frac{x^2}{2}\cdot 1\cdot 0
\end{align*}$$We can $(1)$ and $(2)$ because both limits exist and are finite.
Unless $x=0$ which can be simply plugged to the original limit as $\cos 0=1$
For $m$ near $\infty$,
$$\cos(\frac xm)=1-\frac{x^2}{2m^2}(1+\epsilon(m))$$
$$\ln(\cos(\frac xm))=-\frac{x^2}{2m^2}(1+\epsilon(m))$$
$$(\cos(\frac xm))^m=e^{m\ln(\cos(\frac xm))}=$$$$\Large{e^{-\frac{x^2}{2m}(1+\epsilon(m))}}$$
So, $$\lim_{|m|\to+\infty}(\cos(\frac xm))^m=1$$
$\endgroup$ $\begingroup$After almost 2 weeks, and with your help, I managed to find the obvious solution to problem 1)
$$ \lim_{m\to\infty}\left(\cos\left(\frac{x}{m}\right)\right)^m = \lim_{m\to\infty}\left[\cos^2\left(\frac{x}{m}\right)\right]^{m/2} = \lim_{m\to\infty}\left[1-\sin^2\left(\frac{x}{m}\right)\right]^{m/2} = \lim_{m\to\infty}\left[1-\left(\sin\left(\frac{x}{m}\right)/\frac{x}{m}\right)^2\frac{x^2}{m^2}\right]^{m/2} = \lim_{m\to\infty}\left[1-\lim_{m\to\infty}\left(\sin\left(\frac{x}{m}\right)/\frac{x}{m}\right)^2\frac{x^2}{m^2}\right]^{m/2} = \lim_{m\to \infty}\left[1-\lim_{w\to 0}\left(\sin\left(w\right)/w\right)^2\frac{x^2}{m^2}\right]^{m/2} = \lim_{m\to \infty}\left[1-1\times\frac{x^2}{m^2}\right]^{m/2} = \lim_{m\to \infty}\left[\left(1-\frac{x}{m}\right)\left(1+\frac{x}{m}\right)\right]^{m/2} = \lim_{m\to \infty}\left(1-\frac{x}{m}\right)^{m/2} \times \lim_{m\to \infty}\left(1+\frac{x}{m}\right)^{m/2} = \lim_{y\to \infty}\left(1+\frac{1}{y}\right)^{-xy/2} \times \lim_{m\to \infty}\left(1+\frac{1}{z}\right)^{xz/2} $$
Thanks everyone for your help
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