Question

Limits. In this problem why was it necessary to take the ln. & how do you know when to do so?

Answer 1

Good morning, child.

My proposition is that logarithms are utilised whenever you need the exponent down. In that case, you take the logarithm of both sides.

May He bless you.

Answer 2

Was the reply prior to this one able to assist you?

Answer 3

you do it here so that you manipulate the problem and take the limit in another form

Answer 4

Child?

Answer 5

to use logarithm is just a way to change an expretion so you have no exponents arround and later use L'Hopittal rule. This is a trick that you can learn to use just by experience, there is no general reason for it

Answer 6

@myko would you mind helping with a similar problem?

Answer 7

\[\lim_{x \rightarrow \infty}\left( 1+\frac{ 3 }{ x } \right)^{-x}\]

Answer 8

i worked it out and got the answer of one

Answer 9

\[\ln(y) = -x \ln \left(\dfrac{x + 3}{x}\right)\]\(x \to \infty\).

Answer 10

\[\ln(y) = -\infty \]\[y = e^{-\infty} = 0.\]

Answer 11

@elaornelas, there are at least two good ways to approach this limit
\[\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{-x}\]
Recall the limit definition of \(e\):
\[e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\]
This first method involves rewriting the limit so that it can be expressed in this exponential form.

Let's substitute, \(n=\dfrac{x}{3}\), then using some properties of limits and continuity, we have
\[\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{-x}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{-3n}=\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\right)^{-3}=e^{-3}\]
The other limit, if you're not comfortable with the substitution method, would involve logarithms. What we do is assume the limit exists, call it \(L\).
\[L=\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{-x}\]
Take the log of both sides:
\[\ln L=\ln\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{-x}\]
The limit expression is continuous as \(n\to\infty\), which allows you to rewrite as
\[\ln L=\lim_{x\to\infty}\ln\left(1+\frac{3}{x}\right)^{-x}\]
Log properties:
\[\ln L=\lim_{x\to\infty}-x\ln\left(1+\frac{3}{x}\right)\]
Direct substitution yields an indeterminate form, \(-\infty\cdot0\), so you must rewrite before you can apply L'Hopital's rule:
\[\ln L=-\lim_{x\to\infty}\frac{\ln\left(1+\dfrac{3}{x}\right)}{\dfrac{1}{x}}=\frac{0}{0}\]
No apply L'Hopital's rule:
\[\ln L=-\lim_{x\to\infty}\frac{-\dfrac{\dfrac{3}{x^2}}{1+\dfrac{3}{x}}}{-\dfrac{1}{x^2}}=-\lim_{x\to\infty}\dfrac{3}{1+\dfrac{3}{x}}=-3\]
Then taking the exponential of both sides,
\[e^{\ln L}=e^{-3}~~\iff~~L=e^{-3}\]