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Find the value for the constant $k$ such that

$$\lim_{x\rightarrow\infty} \frac{(4^{kx}+6)}{(4^{2x}+4)}$$


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Can you guess?? – Ross Millikan Jan 15 '13 at 21:25
Hint: What happens when $k > 2$? – mrf Jan 15 '13 at 21:25
Hi! Welcome to math.SE! This forum is full of free-spirited people who are happy to respond to questions. You have techincally written a question, but lots of potential answerers are going to go help another poster who frames his/her question nicely by including partial work and thoughts, and who phrased their question less like a command. You might consider making such changes to improve your chances of getting a good answer. – rschwieb Jan 15 '13 at 21:28
up vote 3 down vote accepted

Divide all terms in numerator and denominator by $4^{2x}:$ $$L= \lim_{x\rightarrow +\infty} \frac{4^{kx}+6}{4^{2x}+4}= \lim_{x\rightarrow+\infty} \frac{4^{(k-2)x}+\dfrac{6}{4^{2x}}}{1+\dfrac{4}{4^{2x}}}$$ Since $\dfrac{6}{4^{2x}}\underset{x\to{+\infty}}{\to} 0$ and $\dfrac{4}{4^{2x}}\underset{x\to{+\infty}}{\to} 0$ then depending on value of $k$ we have following possibilities:

  1. $k>2 \Rightarrow {4^{(k-2)x} \underset{x\to{+\infty}}{\to} {+\infty}},$ therefore, $L=+\infty$;
  2. $k=2 \Rightarrow L=1 $;
  3. $k<2 \Rightarrow L=0 $.
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so would this leave me with 1^(k-2)? Its been a while since Ive had to divide exponents which leaves me very rusty – user57172 Jan 15 '13 at 21:32
oh so k would have to be equal to or less than 2 in order for the limit to exist? – user57172 Jan 15 '13 at 21:50
Yes, limit exist for $k\leqslant{2}.$ – M. Strochyk Jan 15 '13 at 21:54

Hint: when $x$ gets large, the $6$ is negligible compared with $4^{kx}$. You can make this explicit by dividing through by $4^{kx}$

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so would that leave me with 1/1^(2-k) or did I do that wrong? – user57172 Jan 15 '13 at 21:35
@user57172: you are right, you can ignore the $+4$ as well. No, it should be $\frac 1{4^{(2-k)x}}$. Now can you find what happens for various values of $k$ as $x$ gets large? – Ross Millikan Jan 15 '13 at 21:39
wouldnt the limit go towards infinity for any number of k though, other than 2 itself? – user57172 Jan 15 '13 at 21:46
@user57172: No, if $k=1$, for example, it will decrease to zero. – Ross Millikan Jan 15 '13 at 21:50
oh ok I see that now! – user57172 Jan 15 '13 at 21:53

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