Need help understanding how these limits were evaluated

[kostoglotov]

Homework Statement

Hi, the problem is imply to show the following

$$\lim_{n\rightarrow \infty} 10^n e^{-t} \sinh{10^{-n}t} = \lim_{n\rightarrow \infty} 10^n e^{-t} \sin{10^{-n}t} = te^{-t}$$

How can I do this? Just a hint or a first step would be great, thanks :)

Homework Equations

The Attempt at a Solution

I've tried to make the substitution y = ln[f(x)], but I didn't get far with that.

 

[Mark44]

Homework Statement

Hi, the problem is imply to show the following

$$\lim_{n\rightarrow \infty} 10^n e^{-t} \sinh{10^{-n}t} = \lim_{n\rightarrow \infty} 10^n e^{-t} \sin{10^{-n}t} = te^{-t}$$

How can I do this? Just a hint or a first step would be great, thanks :)

Homework Equations

The Attempt at a Solution

I've tried to make the substitution y = ln[f(x)], but I didn't get far with that.

I'm pretty sure they used L'Hopital's Rule after bringing out $e^{-t}$.

 

[BvU]

No typos ? Then you can just pull out the $\ \ t\; e^{-t}$ !
Do you know the $$\lim_{ y\downarrow 0}\ {\sin y\over y }\quad {\rm ?} $$

 

[kostoglotov]

No typos ? Then you can just pull out the $\ \ t\; e^{-t}$ !
Do you know the $$\lim_{ y\downarrow 0}\ {\sin y\over y }\quad {\rm ?} $$

That's what I ended up using :) But I'm not sure if it's ok to make the substitution n = 1/a whereby as n -> inf, then a -> 0...

 

[kostoglotov]

No typos ? Then you can just pull out the $\ \ t\; e^{-t}$ !
Do you know the $$\lim_{ y\downarrow 0}\ {\sin y\over y }\quad {\rm ?} $$

I also don't understand why it is works for the hyperbolic sine, but I'm happy to just accept that it does, and maybe explore the proof that it does separately; this problem isn't part of a precalculus or limits specific study effort, it appeared as part of a challenge problem in a chapter of a differential equations text I'm working on.

 

[BvU]

For $\sinh x \equiv {e^x-e^{-x}\over 2} \ \Rightarrow \ \sin\; i{\bf z} = i \sinh {\bf z}$
or you can use the Taylor series of $e^x = 1 + x + {1\over 2}x^2 + ...$ to show that
$$\lim_{ y\downarrow 0}\; {\sinh y\over y }\quad = 1 $$

 

[Ray Vickson]

I also don't understand why it is works for the hyperbolic sine, but I'm happy to just accept that it does, and maybe explore the proof that it does separately; this problem isn't part of a precalculus or limits specific study effort, it appeared as part of a challenge problem in a chapter of a differential equations text I'm working on.

It is much simpler than the case of $\sin(x)/x$. Using the definition of $\sinh(x)$ we have
$$\begin{array}{rcl}\sinh(x) &=&\frac{1}{2}(e^x - e^{-x}) \\ &=& \frac{1}{2} [1 + x + x^2/2 + x^3/6 \cdots - (1 - x + x^2/2 - x^3/6 + \cdots)]\\ &=& x + x^3/6 + \cdots \end{array}$$
so $\sinh(x)/x = 1 + x^2/6 + \cdots \to 1$ as $x \to 0$. Alternately, just use l'Hospital's rule on $f(x) = \sinh(x)/x$.