# Thread: Problem #7 - Limit

1. ## Problem #7 - Limit

This one is from our very own archives. Just for the record you can find it here. But what kind of challenge would it be to simply apply l'Hopital's rule? So....

Find the limit without using l'Hopital's rule or power series.

$\lim_{x \to \pi ^+} (1 + 3 sin(x) )^{cot(x)}$

-Dan

2. ## Re: Problem #7 - Limit

This is one of those "you have to see it before you can do it" problems. Generally if I see a limit where the exponent has a variable in it then I try out the "exponential limit":
$\lim_{u \to \infty} \left ( 1 + \frac{1}{u} \right ) ^u = e$

So let's give it a try. If we set $u = \frac{1}{3~sin (x)}$ then we know that
$sin(x) = \frac{1}{3u}$

$cos(x) = -\sqrt{1 - sin^2(x)} = -\sqrt{1 - \frac{1}{9u^2}}$
(x is in the second quadrant, hence the negative sign.)

and thus
$cot(x) = \frac{cos(x)}{sin(x)} = -\frac{\sqrt{1 - \frac{1}{9u^2}}}{\frac{1}{3u}} = -3u \sqrt{1 - \frac{1}{9u^2}}$

and the limit now goes as $u \to \infty}$.

So our limit has the form:
$\lim_{x \to \pi ^+} \left ( 1 + 3~sin(x) \right ) ^{cot(x)} = \lim_{u \to \infty} \left ( 1 +\frac{1}{u} \right ) ^{-3u \sqrt{1 - 1/(9u^2)}}$

$= \lim_{u \to \infty} \left ( 1 +\frac{1}{u} \right ) ^{u \left ( -3 \sqrt{1 - 1/(9u^2)} \right )}$

Now we have to be a bit careful on the rigorous side of the street. Note that, before we take the limit, that the exponent is a continuous function for non-zero u. Thus we can use the following principle:
If c(u) = a(u) ^{b(u)}, then $\lim c(u) = \left ( \lim a(u) \right ) ^{ \lim b(u) }$ where a(u) and b(u) are continuous at the limit "point."

Thus
$\lim_{x \to \pi ^+} \left ( 1 + 3~sin(x) \right ) ^{cot(x)} = \lim_{u \to \infty} \left ( 1 +\frac{1}{u} \right ) ^{u \left ( -3 \sqrt{1 - 1/(9u^2)} \right )}$

$= \left [ \lim_{u \to \infty} \left ( 1 + \frac{1}{u} \right ) ^u \right ] ^{ \left [ \lim_{u \to \infty} -3 \sqrt{1 - 1/(9 u^2)} \right ]$

$= e^{-3} = \frac{1}{e^3}$

-Dan

3. ## Re: Problem #7 - Limit

Limx→π+(1+3sinx)^(cosx/sinx) = limx→0+(1-3x)^(1/x) = limn→∞(1-3/n)n = e-3

If you accept lim (1+x/n)n = ex, and some basic theorems about limits.