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Thread: Prove power of limits using Mathematical induction.

  1. #1
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    Question 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?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Arturo_026 View Post
    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 $\displaystyle n=1$. We know it's true for $\displaystyle n=1$ because that fact is given to us.

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

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

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

    So we're done.
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    Step 1: Show it's true for $\displaystyle n=1$. We know it's true for $\displaystyle n=1$ because that fact is given to us.

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

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

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

    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.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Arturo_026 View Post
    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.

    $\displaystyle \lim_{x\to a}f(x)$ is written as \lim_{x\to a}f(x). $\displaystyle \int_a^b f(x)\,dx$ is written as \int_{a}^{b}f(x)dx. $\displaystyle \frac{a}{b}$ is written as \frac{a}{b}. $\displaystyle \sqrt{a}$ is \sqrt{a}. $\displaystyle a^b$ 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 ($\displaystyle \Sigma$) button on the toolbar.
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  5. #5
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    Re: Prove power of limits using Mathematical induction.

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

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

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

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

    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
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  6. #6
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    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.

    Quote Originally Posted by redsoxfan325 View Post
    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.
    Last edited by SlipEternal; Jan 10th 2018 at 06:37 AM.
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