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Here's the exercise and my work: http://imgur.com/YFBLOHb

The result is correct, but I'm not sure if I got to it in a strictly correct way.

Do I have to do the same process for negative infinity? 1/x could approach negative infinity but the limit for negative infinity should be the same if the original problem actually has a solution.

Am I allowed to move the x to the other side of the equation between line 3 and 4? It approaches infinity, and moving infinity like that wouldn't be allowed, but I'm not sure if it's the case when the x is still in limits.

Similar for dividing by x and then later putting everything to the xth power, is either of them not allowed?

And finally, this one is unrelated, is this: http://imgur.com/gI3hlmL true? Why?

Thanks in advance.

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    Your method is flawed unfortunately. – 2017-01-27
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    Unless you type in the math instead of relying on those poor images, I think this question will be closed. Those images can easily disappear in time. – 2017-01-27
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    If you split a limit, you have to know that both limits exist. When you pull the x off to the other side, you break it. I think you finding the right answer was because you knew what to expect, because there are other possible answers you can draw when you break limit laws. – 2017-01-27

2 Answers 2

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If we are allowed to use equivalent infinitesimals and taking into account that $a^x\sim x\log a$ as $x\to 0$ then, $$L=\lim_{x\to 0}\frac{6^x-1}{x}=\lim_{x\to 0}\frac{x\log 6}{x}=\lim_{x\to 0}\log 6=\log 6.$$

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    How do you know $a^x\sim x\log a$ as $x\to 0$ without prior Calculus? To me that looks like a tangent to the exponential curve at zero and that requires a derivative or something? – 2017-01-27
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    I don't know. For that reason I said "If we are allowed". Sometimes, teahers provide theoric results whithout proof. (+1 for your answer) – 2017-01-28
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What makes me a little "unsettled" when I read your work is that you are using couple of limit laws, like the limit of a sum is the sum of the limits. Problem is that one of your limits goes to infinity and so great care needs to be taken with those limit laws, even if you got the right answer. But why not use one of the definitions of the derivative? $\frac{6^x-6^0}{x-0}$ really is the derivative of the function $y=6^x$, taken at $0$. So plug in $x=0$ in $6^xln6$ to arrive at your answer.

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    I know it can be solved using derivatives, but we haven't done those at school yet and this problem was supposed to be solved only using fairly limited limit knowledge. I know this can be problematic, that's why I'm asking. – 2017-01-27
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    @TAJD I am hopeful to see how your teacher solves this problem without derivatives, or LHospital's Rule, or the answer provided by Fernando...Because that all requires derivatives at some point. – 2017-01-27