2

If $a,b,c,d$ are real constants such that $$\lim_{x\to 0}\dfrac{ax^2+\sin bx+\sin cx +\sin dx}{3x^2+5x^4+7x^6}=8$$ Then find $a+b+c+d$.

Answer: $24$

I made two different attempts to solve the above question, of which one got me the right answer while the other didn't get me an answer actually had a typo which fixes the solution.

METHOD $1$: (That works)

Since $x\to 0\Rightarrow bx,\ cx,\ dx \to 0$. So the limit becomes $$\lim_{x\to 0}\frac{ax^2 +(b+c+d)x}{3x^2+5x^4+7x^6}$$ Using $\lim_\limits{x\to 0} \frac{\sin x}{x}=1$.

Now, the coefficient of $x$ must be zero for the finite limit to exist i.e. $$b+c+d=0\Rightarrow a+b+c+d=a$$ Also, the limit transforms to $$\lim_{x\to 0}\frac{ax^2}{3x^2+5x^4+7x^6}=\frac a3=24$$ So $a=24$ and so is the required sum.

METHOD $2$: (That initially did not work but now does)

Applying the Taylor Series expansion, $$\sin x=x-\frac {x^3}{3!}+ O(x^5)$$ The limit becomes, $$\lim_{x\to 0}\frac{(b+c+d)x+ax^2+\left(\dfrac{b^3+c^3+d^3}{6}\right)x^3+O(x^5)}{3x^2+5x^4+7x^6}$$

From here, one can now conclude in a fashion similar to Method $1$.

(Edit: Several users have pointed in the comments that my Taylor Series was actually that of $\cos x$. That was a typo and it solves my question. I have edited the Method $2$ so that now it is correct. )

My Question: Why does the second method fail? Also, how to determine whether a question wants the first or the second method(or any other method)?

Thank you in advance.

Edit 2: After a lengthy discussion in chats with @Paramanand Singh, I have now understood that Method $1$ is wrong.

AryanSonwatikar
  • 2,728

1 Answers1

1

Your first approach is wrong. Don't get discouraged as your mistake is very common among beginners. Instead of $\lim_{x\to 0}\dfrac{\sin x} {x} =1$ the limit needed here is $$\lim_{x\to 0}\frac{\sin (kx) -kx} {x^2}=0\tag{1}$$ for any constant $k$. This is obvious if $k=0$ and not that difficult if $k\neq 0$. Now replace the denominator by $3x^2$ and using $(1)$ the hypotheses of the question lead to $$\lim_{x\to 0}\frac{ax^2+(b+c+d)x}{3x^2}=8\tag{2}$$ and this leads to $$\lim_{x\to 0}\frac{b+c+d}{3x}=8-\frac{a}{3}$$ This gives us $b+c+d=0$ and $a=24$ as in your approach.

The limit $\lim_{x\to 0}\dfrac{\sin x} {x} =1$ can not really give you equation $(2)$.

Paramanand Singh
  • 87,309
  • Actually I think it can: we can write $\sin kx$ as $\frac{\sin kx}{kx} \cdot kx$. If we tend $x$ to $0$ with $k$ being finite, $kx$ tends to $0$ and now we can apply the limit $\lim_{kx\to 0} \frac{\sin kx}{kx}=1$. And we can replace $\sin kx$ by $kx$. Please correct me if I'm wrong. – AryanSonwatikar May 18 '20 at 09:31
  • @AryanSonwatikar: the fact that $(\sin x) /x\to 1$ as $x\to 0$ does not mean that you can replace $\sin x$ by $x$ while evaluating limits as $x\to 0$. Rather it means that you can replace the expression $\lim\limits_{x\to 0}\dfrac{\sin x} {x}$ with $1$. You can convince yourself by changing $\sin x$ with $\log(1+x)$ in your problem. Not that $(\log(1+x))/ x\to 1$. Doing this changes the problem considerably. – Paramanand Singh May 18 '20 at 09:50
  • @AryanSonwatikar: I am surprised how beginners think of limit formulas as justification for algebraic replacement of one expression by another expression. No calculus book tell this and it is an imagination of students. – Paramanand Singh May 18 '20 at 09:52
  • 1
    You are wrong in the reasoing. Not in the result, as (1) shows. You are not wrong "by much" (your argument can be expanded to work). But as stated, it's wrong in general. It's tempting, in a a limit, to substitute part of a term with "the limit" (as you did by substituting $kx$ for $\sin(kx)$), but that can lead to wrong results. The simplest example is $\lim_{n\to \infty} (1+\frac1n)^n$, which equals $e$. But using your "technique", it would be possible to argue that $\lim_{n\to \infty} (1+\frac1n)=1$ (correct) and conclude that $\lim_{n\to \infty} (1+\frac1n)^n = \lim_{n\to \infty} 1^n=1$ – Ingix May 18 '20 at 09:54
  • @Ingix: good examples. +1 for your comment. – Paramanand Singh May 18 '20 at 09:55
  • @AryanSonwatikar: also you may ask how come the replacement of denominator by $3x^2$ is valid. Well that is obtained by multiplying given limit in question with another limit $$\lim_{x\to 0}\frac{3x^2+5x^4+7x^6}{3x^2}=1$$ since both limit exist you can multiply and get another result. – Paramanand Singh May 18 '20 at 09:58
  • $\require{cancel}$ @ingix I understand your points. I actually used "replace" in the informal sense. Let me rephrase. $\lim_{x\to 0} \sin kx= \lim_{x\to 0} \cancelto {1}{\frac{\sin kx}{kx}}\cdot kx$. Is this valid? – AryanSonwatikar May 18 '20 at 10:08
  • @AryanSonwatikar: this is valid and both sides of equation are $0$. From this you can't conclude replacement of $\sin kx$ by $kx$. I really don't see how that idea of replacement comes into picture. – Paramanand Singh May 18 '20 at 10:33
  • @Parmanand Singh "replacement" was used informally. The above is what I have done in Method $1$ without writing it in such detail. – AryanSonwatikar May 18 '20 at 10:35
  • @AryanSonwatikar: Perhaps this is what you think. Since $\lim_{x\to 0}\sin kx=\lim_{x\to 0}kx$ we can as well remove limit operator and think of $\sin kx$ being almost like $kx$ and thus replace. This line of thinking has to be ditched seriously if you want to become expert in calculus. – Paramanand Singh May 18 '20 at 10:37
  • @AryanSonwatikar: let me reiterate. Your method 1 is wrong because it is based on a wrong idea. You will not get credit for this solution if it is evaluated by an expert examiner. You should have tried replacing $\sin x$ with $\log(1+x)$ in the problem and see the failure of your approach in more direct manner. – Paramanand Singh May 18 '20 at 10:39
  • @AryanSonwatikar: adding any details in your approach won't make it correct. The real justification of the approach is via the limit formula $(1)$ of my answer. Let me know if you need more help here. – Paramanand Singh May 18 '20 at 10:40
  • @Paramanand Singh Alright, please forget I said replace. Consider that a brainfart from my side. Can I use what I have done in the comment I posted nearly $30$ minutes ago in Method $1$ to arrive at what I have? – AryanSonwatikar May 18 '20 at 10:41
  • @AryanSonwatikar: no you can't do that. Also don't get me wrong. I am really trying to give you the right ideas. Calculus is made difficult by prevalence of such ideas which actually have no basis. Instead focus on the right concepts and approaches. – Paramanand Singh May 18 '20 at 10:44
  • @AryanSonwatikar: also see an answer to a related question: https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh May 18 '20 at 10:46
  • @Paramanand Singh Consider this answer of yours. In the second step after defining $f(n)$, you write the denominator from $x^2\sin x$ to $x^3$. I have used the same method (according to me) to arrive at what I have arrived at; just that I have used the word "replace" informally. – AryanSonwatikar May 18 '20 at 10:47
  • Let us continue this discussion in chat. – Paramanand Singh May 18 '20 at 10:50