Thread: Prove power of limits using Mathematical induction.

1. Prove power of limits using Mathematical induction.

It has been a while that I've done math. induction and I'm having trouble with this challenge:

Use mathematical induction to prove that if n is a positive integer and lim f(x) as x-->a = L ; the lim [f(x)]^n as x-->a = [L]^n

How should this proof be approached?

2. Originally Posted by Arturo_026
It has been a while that I've done math. induction and I'm having trouble with this challenge:

Use mathematical induction to prove that if n is a positive integer and lim f(x) as x-->a = L ; the lim [f(x)]^n as x-->a = [L]^n

How should this proof be approached?
Step 1: Show it's true for $n=1$. We know it's true for $n=1$ because that fact is given to us.

Step 2: Assume it's true for $n=k$. So assume $\lim_{x\to a}f^k(x)=L^k$.

Step 3: Prove it's true for $n=k+1$. This is shown below:

$\lim_{x\to a}f^{k+1}(x) = \underbrace{\lim_{x\to a}\left(f^k(x)\cdot f(x)\right) = \left(\lim_{x\to a}f^k(x)\right)\left(\lim_{x\to a}f(x)\right)}_{Limit~Product~Rule} = L^k\cdot L = L^{k+1}$

So we're done.

3. Originally Posted by redsoxfan325
Step 1: Show it's true for $n=1$. We know it's true for $n=1$ because that fact is given to us.

Step 2: Assume it's true for $n=k$. So assume $\lim_{x\to a}f^k(x)=L^k$.

Step 3: Prove it's true for $n=k+1$. This is shown below:

$\lim_{x\to a}f^{k+1}(x) = \underbrace{\lim_{x\to a}\left(f^k(x)\cdot f(x)\right) = \left(\lim_{x\to a}f^k(x)\right)\left(\lim_{x\to a}f(x)\right)}_{Limit~Product~Rule} = L^k\cdot L = L^{k+1}$

So we're done.
Wow, i didn't imagine it'll be so short.
Thank you so much.

P.S. How do u get the limit notations in that format, I tried wolfram alpha but can.

4. Originally Posted by Arturo_026
Wow, i didn't imagine it'll be so short.
Thank you so much.

P.S. How do u get the limit notations in that format, I tried wolfram alpha but can.
Click on the math equations in my posts and a window will pop up showing you the code.

$\lim_{x\to a}f(x)$ is written as \lim_{x\to a}f(x). $\int_a^b f(x)\,dx$ is written as \int_{a}^{b}f(x)dx. $\frac{a}{b}$ is written as \frac{a}{b}. $\sqrt{a}$ is \sqrt{a}. $a^b$ is a^{b}. Surround the expressions with math tags. The close math tag is [/tex] and the opening one is [tex]. Or you can just highlight the expression and click the Sigma ( $\Sigma$) button on the toolbar.

5. Re: Prove power of limits using Mathematical induction.

Originally Posted by redsoxfan325
Step 1: Show it's true for $n=1$. We know it's true for $n=1$ because that fact is given to us.

Step 2: Assume it's true for $n=k$. So assume $\lim_{x\to a}f^k(x)=L^k$.

Step 3: Prove it's true for $n=k+1$. This is shown below:

$\lim_{x\to a}f^{k+1}(x) = \underbrace{\lim_{x\to a}\left(f^k(x)\cdot f(x)\right) = \left(\lim_{x\to a}f^k(x)\right)\left(\lim_{x\to a}f(x)\right)}_{Limit~Product~Rule} = L^k\cdot L = L^{k+1}$

So we're done.

why cant i see the solution, sir. i actually got the same question and this would really be a help if i'd be able to get it too. im really clueless to the third step hhu seeking help :c

6. Re: Prove power of limits using Mathematical induction.

Here is the post with the tex tags switched to dollar signs so that the latex renders.

Originally Posted by redsoxfan325
Step 1: Show it's true for $n=1$. We know it's true for $n=1$ because that fact is given to us.

Step 2: Assume it's true for $n=k$. So assume $\displaystyle \lim_{x\to a}f^k(x)=L^k$.

Step 3: Prove it's true for $n=k+1$. This is shown below:

$\displaystyle \lim_{x\to a}f^{k+1}(x) = \underbrace{\lim_{x\to a}\left(f^k(x)\cdot f(x)\right) = \left(\lim_{x\to a}f^k(x)\right)\left(\lim_{x\to a}f(x)\right)}_{\text{Limit Product Rule}} = L^k\cdot L = L^{k+1}$

So we're done.