1

Can someone help me with this problem? I don't know what to do after a certain step...

$$\lim_{x\to \infty} \left(\frac{4x - 3} {4x+5}\right)^{4x+1}$$

So far what I have done is take the natural log of both sides and then that takes the exponent out and brings to the front... so then we use one of the natural log rules to get:

$$ ln y = \lim_{x\to \infty} (4x+1)(({ln (4x - 3)})- (ln(4x+5)))$$

This is where I am stuck and have no idea how to continue, what should I do next?

CloudN9ne
  • 349
  • 1
    One way is to rewrite as $\frac{\ln(4x-3)-\ln(4x+5)}{(1/(4x+1)}$ and use L'Hospital's Rule. another way, if you officially know it, is to use the fact that $\lim_{m\to\infty}\left(1+\frac{x}{n}\right)^n=e^x$. – André Nicolas Feb 13 '14 at 02:16
  • Are you allowed to use L'Hopital's Rule? If so it is not very different from $\lim_{x\to0}x\ln x$. The one thing that you mustn't do (though it might initially look like a good idea) is to split into two limits - then you will get $\infty-\infty$ which is meaningless. – David Feb 13 '14 at 02:18
  • Yes, I am allowed. I'll give it a try. – CloudN9ne Feb 13 '14 at 02:18

1 Answers1

2

$$\lim_{x\to \infty} \left(\frac{4x - 3} {4x+5}\right)^{4x+1} = \lim_{x\to \infty} \left(1 -\frac{8} {4x+5}\right)^{\frac{(4x+1)(4x+5)}{4x+5}} = \lim_{x\to \infty} \left(\left(1 -\frac{8} {4x+5}\right)^{4x+5}\right)^{\frac{4+\frac 1x}{4+\frac 5x}} = (e^{-8})^1 = \frac 1{e^8}$$

Stefan4024
  • 35,843