Ln Rule Simplification Question?

[Silvanoshei]

Book has this answer as 1, was wondering how they simplified it down.

\(\displaystyle ln(4x-1)-ln(4x+1) \rightarrow ln\frac{4x-1}{4x+1} x\rightarrow\infty=1\)

Tried taking some stuff out to the side...

\(\displaystyle \frac{4}{4}*\frac{x-1}{x+1} as\rightarrow x\rightarrow \infty=\infty\)

Confused on this, shouldn't this be \(\displaystyle \infty\)?

 

[Deleted member 4993]

Book has this answer as 1, was wondering how they simplified it down.

\(\displaystyle ln(4x-1)-ln(4x+1) \rightarrow ln\frac{4x-1}{4x+1} x\rightarrow\infty=1\)

Tried taking some stuff out to the side...

\(\displaystyle \frac{4}{4}*\frac{x-1}{x+1} as\rightarrow x\rightarrow \infty=\infty\)

Confused on this, shouldn't this be \(\displaystyle \infty\)?

\(\displaystyle \displaystyle{\lim_{x \to \infty}\left[ln(4x-1) - ln(4x+1)\right]}\)

\(\displaystyle = \ \displaystyle{\lim_{x \to \infty}ln\left[\frac{4x-1}{4x+1}\right]}\)

\(\displaystyle = \ \displaystyle{\lim_{x \to \infty}ln\left[\frac{4-\frac{1}{x}}{4+\frac{1}{x}}\right]}\)

\(\displaystyle = \ \displaystyle{ ln\left[\frac{4}{4}\right]}\)

= 0

 

[Silvanoshei]

Wait, you pulled the x out of the num. and dem., but doesn't that mean the x is still going to infinity and multiplying by ln(4/4)?