1

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?

snulty
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Ekanshdeep Gupta
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3 Answers3

1

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

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.

Elliot
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0

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

joefu
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  • That limit would become 0/x^2. And, as demonstated here: http://math.stackexchange.com/questions/867967/why-limit-of-0-x-is-0-if-x-approaches-0, limit of 0/x^2 = 0 when x approaches to zero. – Ekanshdeep Gupta Jun 25 '16 at 19:43
  • 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. – imranfat Jun 25 '16 at 19:59
  • 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. – joefu Jun 25 '16 at 20:01
  • @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 $\lim g(x)=\infty$. – symplectomorphic Jun 25 '16 at 20:02
  • Alright, got it... Thanks everyone. – Ekanshdeep Gupta Jun 25 '16 at 20:10
0

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.

Community
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Paramanand Singh
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  • Thanks a lot, that helped! Can you send a link to your mentioned blog, where I might read these rules more fully? – Ekanshdeep Gupta Jun 26 '16 at 16:23
  • @EkanshdeepGupta: please have a look at http://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-1.html – Paramanand Singh Jun 26 '16 at 17:36
  • 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? – Ekanshdeep Gupta Jun 28 '16 at 19:10
  • @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. – Paramanand Singh Jun 28 '16 at 19:27