# Thread: Quick question on a limit

1. ## Quick question on a limit

Hi, Im having a little trouble understanding a limit. The limit as k goes to infinity of k/(k+1), my book says the answer is 1. I tried to divide everything by k and got 1/infinity = 0? Can someone please try and explain how this limit is equal to 1.

Thank you

2. Hello,

This is a very classical one

$\displaystyle \frac{x}{x+1}$

Divide above and below by x powered to the biggest coefficient encountered.
So here, divide by x.

$\displaystyle =\frac{\frac{x}{x}}{\frac{x}{x}+\frac{1}{x}}=\frac {1}{1+\frac{1}{x}}$

But if any constant number is divided by something that tends to infinity, the resulting limit will be 0.

So $\displaystyle \lim_{x \to + \infty} \frac{1}{1+\frac{1}{x}}=\frac{1}{1}=1$

3. Hello

Dividing by $\displaystyle k$ is a good idea, it yields $\displaystyle \frac{k}{k+1}=\frac{1}{1+\frac{1}{k}}$
As $\displaystyle \lim_{k\to\infty}\frac{1}{k}=0$, $\displaystyle \lim_{k\to\infty}\frac{1}{1+\frac{1}{k}}=1$

4. thanks for the help

5. Or just make $\displaystyle \frac{k} {{k + 1}} = \frac{{k + 1 - 1}} {{k + 1}} = 1 - \frac{1} {{k + 1}}$ and as $\displaystyle k\to\infty$ the answer is pretty obvious.

6. Originally Posted by Moo
Hello,

This is a very classical one

$\displaystyle \frac{x}{x+1}$

Divide above and below by x powered to the biggest coefficient encountered.
So here, divide by x.

$\displaystyle =\frac{\frac{x}{x}}{\frac{x}{x}+\frac{1}{x}}=\frac {1}{1+\frac{1}{x}}$

But if any constant number is divided by something that tends to infinity, the resulting limit will be 0.

So $\displaystyle \lim_{x \to + \infty} \frac{1}{1+\frac{1}{x}}=\frac{1}{1}=1$
O yeah...this one technique Moo taught me is amazing $\displaystyle \lim_{k \to {\infty}}\frac{k}{k+1}=\lim_{k \to {\infty}}\frac{1+0}{1}=1$...its called L'hopital's rule