1. embarrasing algebra question

Hi,

In my calculus book i have the following limit problem:

$\displaystyle \lim_{x\to \infty}~\bigg(1+sin\frac{3}{x}\bigg)^x$

The first part of the solution:

1.Take the log of both sides:

$\displaystyle ln~y = \lim_{x\to \infty}~x~ln\bigg(1+sin\frac{3}{x}\bigg)$

Here's the part i don't understand:

$\displaystyle \frac{ln\bigg(1+sin\frac{3}{x}\bigg)}{\frac{1}{x}}$

Isn't dividing by 1/x the same as multiplying by x?

2. Originally Posted by Jones
Hi,

In my calculus book i have the following limit problem:

$\displaystyle \lim_{x\to \infty}~\bigg(1+sin\frac{3}{x}\bigg)^x$

The first part of the solution:

1.Take the log of both sides:

$\displaystyle ln~y = \lim_{x\to \infty}~x~ln\bigg(1+sin\frac{3}{x}\bigg)$

Here's the part i don't understand:

$\displaystyle \frac{ln\bigg(1+sin\frac{3}{x}\bigg)}{\frac{1}{x}}$

Isn't dividing by 1/x the same as multiplying by x?

Dear Jones,

Here the numerator and the denominator had been divided by x,

$\displaystyle xln\bigg(1+sin\frac{3}{x}\bigg)=\frac{xln\bigg(1+s in\frac{3}{x}\bigg)}{1}=\frac{ln\bigg(1+sin\frac{3 }{x}\bigg)}{\frac{1}{x}}$

3. Originally Posted by Jones
Hi,

In my calculus book i have the following limit problem:

$\displaystyle \lim_{x\to \infty}~\bigg(1+sin\frac{3}{x}\bigg)^x$

The first part of the solution:

1.Take the log of both sides:

$\displaystyle ln~y = \lim_{x\to \infty}~x~ln\bigg(1+sin\frac{3}{x}\bigg)$

Here's the part i don't understand:

$\displaystyle \frac{ln\bigg(1+sin\frac{3}{x}\bigg)}{\frac{1}{x}}$

Isn't dividing by 1/x the same as multiplying by x?

Yes, it is- that the reason you can do this. Now, if you let y= 1/x, that becomes
$\displaystyle \lim_{y\to 0} \frac{ln(1+ sin(3y))}{y}$
That will be of the form "0/0" and you can use L'Hopital's rule.

4. Let y = (x/3) / (3/4). Y can be re-written as (x/3) * (4/3). Whenever you've got a rational expression, you can always multiply the numerator by the reciprocal of the denominator.

(sqrt(x)/2) / 2 = (sqrt(x)/2) * (1/2)