0
$\begingroup$

I've got some limit to show.

$$\lim_{x \rightarrow \infty} \frac{\sum_{n=0}^{\infty}( \frac{x^n}{n!})-1}{1-\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}}$$ What is equivalent to $$\lim_{x \rightarrow \infty} \frac{exp(x)-1}{1-cos(x)}$$

I tried to split it into "even" part and "odd" part, i mean first calculate for $n=0,2,4,6\dots$ and then for $n=1,3,5,7\dots$ but it all got messy and not really led me to any solution. This is my first this kind of task i have to solve, so i don't know any good tricks/ideas i can use here...

I'd appreciate some help! thanks

EDIT: some people state that this limit doesn't exist. But just being curious, what if we want to do series of out it? Seems like it will diverge, right? Why then wolframalpha gives answer = INF to this limit?

$\endgroup$
  • $\begingroup$ What do you mean by "a limit to show"? Do you mean you have to calculate it? $\endgroup$ – 5xum Jan 16 '14 at 13:44
  • $\begingroup$ Sorry for misunderstanding. All time i thought we can "show" inequality/equality/limit/convergance and it is equivalent to "calculating", "solving", if it's not true, then i'm sorry. $\endgroup$ – Krzysztof Lewko Jan 16 '14 at 13:47
  • 1
    $\begingroup$ The denominator is bounded, the numerator isn't. Looks like there is no limit $\endgroup$ – Bernd Jan 16 '14 at 13:47
  • $\begingroup$ @Bernd, if i'm right then $cos(x)$ is bounded by -1 and 1, right? But can we just say that $cos(x)$ is bounded, $exp(x)$ is not, so limit is infinity? $\endgroup$ – Krzysztof Lewko Jan 16 '14 at 13:49
  • $\begingroup$ No you cannot. For example, the limit of $e^x/\cos x$ is NOT infinity, as the function diverges wildly and hits both negative and positive values on any interval $[a, \infty)$. The fact that $1-\cos x \geq 0$ is also important. $\endgroup$ – 5xum Jan 16 '14 at 13:53
1
$\begingroup$

Since $2\geq 1-cos(x) \geq 0$ for all $x$, we have

$$\frac{e^x - 1}{1 - cos(x)} > \frac{e^x - 1}{2}$$

This value becomes arbitrarly large as $x$ becomes large. This means the limit, if any, is $\infty$ (this is of course a generalized limit, no standard limit exists)

Note, however, that even this generalized limit does not actually exist as the function is not defined for all $x$ of the type $x=2\pi k$ for $k\in\mathbb{N}$.

$\endgroup$
  • $\begingroup$ But that just means there is no limit - it doesn't exist. Becuase for the sub sequence you mention the rule for "limit is infinity" does not apply. $\endgroup$ – Bernd Jan 16 '14 at 14:01
  • $\begingroup$ True, but I feel it is in a way "closer" to having a limit than for example $e^x/\cos(x)$. In a way, the function here becomes arbitrarly large for all $x$ for which it is defined. It does not have a limit only because the concept of a limit only exists for functions defined on $[a, \infty)$ for some $a$. The function $e^x/\cos(x)$ does not have a limit for the same reason, but also because it jumps up and down wildly. $\endgroup$ – 5xum Jan 16 '14 at 14:06
  • $\begingroup$ I did second thinking on this task and it seems to be clearer for me now. Thanks! $\endgroup$ – Krzysztof Lewko Jan 21 '14 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.