# Limits Problem

1. Feb 10, 2009

### temaire

1. The problem statement, all variables and given/known data
$$lim_{x\rightarrow\infty}(\frac{3}{5})^{x}$$

2. The attempt at a solution
I know the answer is 0 by using the calculator, but how do I solve this algebraically?

2. Feb 10, 2009

### Dick

Take the log. It's x*log(3/5). As x->infinity what does that approach?

3. Feb 10, 2009

### temaire

I don't see how your method works. If you just add the log, you change the function.

I also have one more.

$$lim_{x\rightarrow\infty}\frac{\sqrt{3x^{2}+x}-\sqrt{x^{2}+2x}}{x}$$

The answer is $$\sqrt{3}-1$$ but I get $$\frac{\sqrt{3}-1}{0}$$

Last edited: Feb 11, 2009
4. Feb 11, 2009

### praharmitra

how do u get a zero???? the limit is towards infinity. Try dividing num and denominator with x. Then apply the limits....

for the first question.... the algebraic way...???? for any number a < 1 limit (x->inf) a^x = 0 always. and 3/5 < 1.

5. Feb 11, 2009

### temaire

If I divide the numerator and denominator by x, wouldn't I end where I started?

But Dick said that you can use logarithms. Can someone show me how to solve my first question using logs? Dick's method was a little confusing to me.

6. Feb 11, 2009

### HallsofIvy

Staff Emeritus
The definition of "continuous function" is that lim f(x)= f(lim x). Logarithm is a continuous function so $lim log((3/5)^x)= log(lim (3/5)^x)$. The lefthand side is x log(3/5). Since 3/5< 1, log(3/5) is negative and x goes to infinity, x log(3/5) goes to negative infinity. In order that log(A)= negative infinity, A must equal 0. $log(lim (3/5)^x))$= negative infinity so $lim (3/5)^x= 0$

Myself, I would consider it simpler, and perfectly valid to note that since 3/5< 1, $3/5)^2< 3/5$, $(3/5)^3< (3/5)^2$ etc. so the limit is 0, as praharmitra said.

7. Feb 11, 2009

### Staff: Mentor

No, not at all. Another way to say this is to factor x from the numerator and denominator. Since lim x/x = 1, as x --> infinity, what you're left with has the same limit.

Getting a factor of x out of the numerator entails taking a factor of x^2 out of each term in each radical, and this factor of x^2 comes out of the radical as a factor of x.