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I check many questions about limit calculations wihtout L'Hopital rule or series representation in this website. Nearly all answers (and best answers) find the limit values without saying "if there is a limit, then it needs to be this $L$".

"Some limit questions which are solved with assumption of existence of limit"

My question: if we assume a limit exists and then if we find a number for this limit, how can we ensure that the limit does really exist?

For example, in this limit: $$\lim_{x\to0}\frac{e^x-x-1}{x^2}.$$

I also want to give a counter-example for assuming the existence of a limit won't provide the limit value that we find is the limit: $$\lim\limits_{x\to 0}\frac{x}{|x|}.$$ Here a limit does not exist. But if we assume $$\lim\limits_{x\to 0} f(x)=L$$ where $f(x)=\frac{x}{|x|}$, then we have $$\lim\limits_{x\to 0} f(x)=\lim\limits_{x\to 0} f(-x)=L.$$ Since $$f(x)+f(-x)=0$$ for all $x \ne 0$ we will have $$2L=0,$$ which gives $L=0$.

user376343
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student forever
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    The only way to ensure the limit exists is to prove it. You've given a nice example to show that finding the value of the limit - assuming the limit exists - is not enough. Take the problem "Find $\lim_{x\to 0}(x-\sin x)/x^3$" for example. The answer should never begin like this:

    $$\lim_{x\to 0}(x-\sin x)/x^3 = \lim_{x\to 0}(x-(x-x^3/6+ \cdots))/x^3 \dots.$$

    Why? We don't know the limit exists yet, so we haven't earned the right to use this notation. (You see this all the time on MSE.) Much better to start off

    $$(x-\sin x)/x^3 = (x-(x-x^3/6+ \cdots))/x^3 \dots.$$

     – zhw. Apr 05 '16 at 16:56
  • Thanks for comment. I am not sure that you saw my question about the limit you wrote: link – student forever Apr 05 '16 at 17:27

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