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I have a solution which gives the answer as $1$:

  1. Write $\tan(x)$ as $\sin(x)/\cos(x)$.
  2. Take $\sin(x)$ common, and multiply&divide by $\cos(x)$.
  3. Rewrite $(1-\cos(x))$ as $2\sin^2(x/2)$
  4. Apply limit for $\sin(x)/x$, $\sin^2(x/2)/x^2$.
  5. That leaves me with $1/(2\cos(x))$, which is $1/2$.

That is what I believe to be the right answer: $1/2$.

But, my real question is, why can't I take $1/(x^2)$ common, and rewrite it as $(\tan(x)/x - \sin(x)/x)(1/x^2)$, apply limit to $\tan(x)/x$ and $\sin(x)/x$ and rewrite it as $(1-1)/(x^2)$, which gives me $0$. What is the flaw here? Is there some rule of limits I'm missing here?

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  • $\begingroup$ While $\tan x/x-\sin x/x$ does tend to 0 as $x\to 0$, $1/x^2$ tends to infinity, so you cannot conclude the product tends to 0. $\endgroup$ Jun 25, 2016 at 19:58
  • $\begingroup$ Another easy way to calculate this limit is to use Maclaurin series for sine and tangent. The numerator is $1/2x^3+O(x^5)$, so the quotient is $1/2+O(x^2)$. $\endgroup$ Jun 25, 2016 at 20:07
  • $\begingroup$ It would look much better if you wrote your initial solution as a sequence of step by step mathematical equations. +1 for raising a good point BTW. $\endgroup$
    – Paramanand Singh
    Jun 26, 2016 at 12:07
  • $\begingroup$ What does taking something "common" mean? $\endgroup$
    – T. M.
    Jun 26, 2016 at 12:23

3 Answers 3

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$$\lim_{x\to a}f(x)g(x)=\left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a}g(x)\right)$$

only if both $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist. The flaw is that $\lim_{x\to 0}x^{-2}=\infty$ so you can not apply the limit law. $1/2$ is the correct answer.

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Note that $(\frac1{x^2})\to \infty \text{ as } x\to 0$

Now $(\frac{\tan x}x - \frac{\sin x}x)\cdot \frac1{x^2}$ is a type of $0\cdot \infty$

which is not clear for us to say something about the convergency of the limit

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  • $\begingroup$ That limit would become 0/x^2. And, as demonstated here: math.stackexchange.com/questions/867967/…, limit of 0/x^2 = 0 when x approaches to zero. $\endgroup$ Jun 25, 2016 at 19:43
  • $\begingroup$ No it doesn't. You have taken the limit of the first part, but you have not taken the limit of the second part, being $\frac{1}{x^2}$. You then simplify your fraction (or whatever it is) and then continue to take the limit on the result. This is not how you can get by in case of product of limits. See Elliot's answer. $\endgroup$
    – imranfat
    Jun 25, 2016 at 19:59
  • $\begingroup$ In $\lim_{x\to 0} \frac0x$ ,the numerator "0" is a number. So 0 times something is still 0. When we say $(\frac{\tan x}x - \frac{\sin x}x)\cdot \frac1{x^2} \to 0$, it means that when x gets close to 0, $(\frac{\tan x}x - \frac{\sin x}x)\cdot \frac1{x^2}$ also gets close to 0, but it do not equal to 0. $\endgroup$
    – joefu
    Jun 25, 2016 at 20:01
  • $\begingroup$ @Ekanshdeep Gupta: no, the link you refer to is irrelevant. Your expression has the form $\lim f(x)\cdot g(x)$. This equals $\big(\lim f(x)\big)\cdot\big(\lim g(x)\big)$ when both limits exist, and you're right $\lim f(x)=0$, but here $\lim g(x)=\infty$. $\endgroup$ Jun 25, 2016 at 20:02
  • $\begingroup$ Alright, got it... Thanks everyone. $\endgroup$ Jun 25, 2016 at 20:10
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It is perfectly valid to write $$\lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}} = \lim_{x \to 0}\left(\frac{\tan x}{x} - \frac{\sin x}{x}\right)\cdot\frac{1}{x^{2}}$$ The next step where you replace the expression in parentheses with $(1-1)$ is invalid and it is invalid precisely because $(1 - 1) = 0$. Had this been non-zero your idea would have been a valid step irrespective of the fact that the remaining part $\lim_{x \to 0}1/x^{2}$ does not exist.

Please see this answer where I explain how and when we can replace a sub-expression by its limit while evaluating the limit of a complex expression in step by step manner.

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  • $\begingroup$ Thanks a lot, that helped! Can you send a link to your mentioned blog, where I might read these rules more fully? $\endgroup$ Jun 26, 2016 at 16:23
  • $\begingroup$ @EkanshdeepGupta: please have a look at paramanands.blogspot.com/2013/11/… $\endgroup$
    – Paramanand Singh
    Jun 26, 2016 at 17:36
  • $\begingroup$ I love the content. It really seems incredibly helpful. However, no math is rendering on your blog. I've used multiple browsers and devices. It makes it harder to read and understand. Decoding equations now takes much more effort. Can that be fixed? $\endgroup$ Jun 28, 2016 at 19:10
  • $\begingroup$ @EkanshdeepGupta: If you are using a mobile device (phone/tablet) then you need to use the browser with desktop mode (this feature is available in chrome / firefox both). If you are using PC/laptop then any browser is fine. However note that math rendering will take some time to load. So you need to be patient. If everything else fails there is a PDF Version available at the end of blog post for each post. You can download that pdf and study. $\endgroup$
    – Paramanand Singh
    Jun 28, 2016 at 19:27

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