# Algebraic Solution to Limits Problem: lim (3/5)^x as x approaches infinity

• temaire
So you're left with lim_{x\rightarrow\infty} x * \frac{\sqrt{3+1/x}-\sqrt{1+2/x}}{x} = lim_{x\rightarrow\infty} \frac{1}{x} * \frac{\sqrt{3+1/x}-\sqrt{1+2/x}}{1}= (0) * \frac{\sqrt{3+0}-\sqrt{1+0}}{1} = 0 * \frac{2}{1} = 0In summary, the limit of (3/5)^x as x approaches infinity is 0. This can be solved algebraically by taking the logarithm, which
temaire

## Homework Statement

$$lim_{x\rightarrow\infty}(\frac{3}{5})^{x}$$

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

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

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:
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.

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

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

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

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.

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.

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

## 1. What is the algebraic solution to this limits problem?

The algebraic solution to the limits problem lim (3/5)^x as x approaches infinity is 0. This can be found by applying the exponential limit rule, which states that for a limit of the form lim a^x as x approaches infinity, the limit is equal to 0 if a is between 0 and 1.

## 2. How do you determine the limit of (3/5)^x as x approaches infinity?

The limit of (3/5)^x as x approaches infinity can be determined by using the exponential limit rule. This rule states that if the base of the exponential expression is between 0 and 1, the limit will be equal to 0. In this case, the base (3/5) is between 0 and 1, so the limit is equal to 0.

## 3. Can you solve this limits problem without using the algebraic solution?

Yes, this limits problem can also be solved using the graphical approach. By graphing the function (3/5)^x, it can be seen that as x approaches infinity, the graph approaches the x-axis, indicating a limit of 0.

## 4. What is the significance of the exponent approaching infinity in this limits problem?

The exponent approaching infinity in this limits problem signifies that the input values are becoming increasingly large. This is important because it allows us to see the behavior of the function at extremely large input values and determine its limit.

## 5. Can the algebraic solution to this limits problem be applied to other exponential expressions?

Yes, the algebraic solution to this limits problem can be applied to other exponential expressions as long as the base is between 0 and 1. This is a general rule for finding the limit of exponential functions as the exponent approaches infinity.

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