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so for solving this question I did the following steps can anyone tell me why my method is wrong. please explain in detail I want to clarify my concept of limits

Question \begin{aligned} &\lim _{x \rightarrow 0}\frac{x^{6000}-(\sin x)^{6000}}{x^{2}(\sin x)^{6000}}\\ \end{aligned}

My solution \begin{aligned} &\lim _{x \rightarrow 0}\left[\left(\frac{x}{\sin x}\right)^{6000} \frac{1}{x^{2}}-\frac{1}{x^{2}}\right]\\ \end{aligned} As we know that \begin{aligned} &\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{6000}=1 \end{aligned} So \begin{aligned} &\lim _{x \rightarrow 0}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}}\right)\ \end{aligned}

so according to me, the answer should be "$0$"

but the answer is $1000$

please explain with a full explanation and it would be better if you give some examples too.

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    $\begingroup$ As for where the argument went wrong: Theorems about "the limit of a product is the product of the limits" require that limits of the individual factors exist. But $\lim\limits_{x\to0}\frac{1}{x^{2}}$ doesn't exist (i.e., is not a real number). $\endgroup$ Apr 15, 2021 at 10:58
  • $\begingroup$ you may not understand the language but can you tell me why the method shown in this video is correct doubtnut.com/question-answer/… $\endgroup$
    – WhyBhav
    Apr 15, 2021 at 11:03
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    $\begingroup$ I can understand the language of the video and advise you stay away from it. Open up your textbook try to learn limit laws and ask questions on limit laws until and unless your understanding of those laws is complete. $\endgroup$
    – Paramanand Singh
    Apr 15, 2021 at 11:18
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    $\begingroup$ The video simply fails to explain why we can replace some expression in denominator with its limit and not do the same in numerator. The thing has nothing to do with $0/0$ or numerator and denominator. Rather it is related to a deep understanding of limit laws. Please go through my linked answer (see next comment) and ask me if something is not clear. $\endgroup$
    – Paramanand Singh
    Apr 15, 2021 at 11:21
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    $\begingroup$ You can't replace a part of an expression with its limit. That's invalid and not permitted by any law of limits. For more details see math.stackexchange.com/a/1783818/72031 $\endgroup$
    – Paramanand Singh
    Apr 15, 2021 at 11:24

4 Answers 4

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I believe the issue comes in where you break up the limit.

Here's the underlying logic of your proof written out more fully:

$$\lim_{x \to 0} \left(\frac{x}{\sin x}\right)^{6000} \frac{1}{x^2} - \frac{1}{x^2} = \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^{6000} \frac{1}{x^2} - \lim_{x \to 0}\frac{1}{x^2} = \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^{6000} \lim_{x \to 0}\frac{1}{x^2} - \lim_{x \to 0}\frac{1}{x^2}$$

$$= (1)\lim_{x \to 0}\frac{1}{x^2} - \lim_{x \to 0}\frac{1}{x^2} = \lim_{x \to 0} \frac{1}{x^2} - \frac{1}{x^2} = 0.$$

Breaking up the limits this way is only valid if all of the limits are finite, which is not true for $\lim_{x \to 0} \frac{1}{x^2}.$

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  • $\begingroup$ in this video youtube.com/… at 10:36 nothing like this is mentioned that splitting needs the limits to be finite , so is the video wrong $\endgroup$
    – WhyBhav
    Apr 15, 2021 at 11:10
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    $\begingroup$ The video is just fine, they mention the rules don't always work. This is one of the conditions where the rules don't work, and they show this with the $\lim_{x \to \infty} x^2 - x$ example. $\endgroup$ Apr 15, 2021 at 11:17
  • $\begingroup$ ok so should i say according to your answer ,splitting of limit is only possible if both the functions are finite and not indeterminate form $\endgroup$
    – WhyBhav
    Apr 15, 2021 at 11:21
  • $\begingroup$ I believe this is the case, yes. By the way, this is why in the Doubtnut video he is able to take the $\left(\frac{\sin x}{x}\right)^{6000}$ out from the denominator in the video: he is breaking the limit over multiplication, which is valid there because if we assume the original limit has a finite value, then both of the limits which he splits into are finite so use of the rule is justified. $\endgroup$ Apr 15, 2021 at 11:26
  • $\begingroup$ are you talking about "youtube" video or "doubtnut" one $\endgroup$
    – WhyBhav
    Apr 15, 2021 at 11:27
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$$\lim_{x\to a} f(x)\cdot g(x)=\lim_{x\to a} f(x)\cdot\lim_{x\to a}g(x)$$ will only work if both the limits are finite in the right hand side

For finite integer $n>0, a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +ab^{n-2}+b^{n-1})\implies$ $$\dfrac{x^n-\sin^nx}{x^2\sin^nx}=\dfrac{x-\sin x}{x^3}\cdot\dfrac x{\sin x}\left(\sum_{r=1}^n\left(\dfrac x{\sin x}\right)^{n-r}\right)$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion and $$\lim_{x\to0}\dfrac{\sin x}x=1$$

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ERROR:

Transforming $$\left(\frac x{\sin x}\right)^{6000} \frac 1{x^2}$$ with $$\lim _{x \to 0}\left(\frac x{\sin x}\right)^{6000}=1$$ into $$\frac 1{x^2}$$ to conclude $$\lim _{x \to 0}\left[\left(\frac x{\sin x}\right)^{6000} \frac 1{x^2}-\frac 1{x^2}\right] \color{red}= \lim _{x \to 0}\left(\frac 1{x^2}-\frac 1{x^2}\right) = \lim 0 = 0$$

makes more or less the same sense as transforming $$\frac xx$$ with $$\lim_{x\to 0}x = 0$$ into $$\frac 0x$$ to conclude $$\lim_{x\to 0} \frac xx \color{red}= \lim \frac 0x = \lim 0 = 0$$

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The biggest problem in your solution is that you have to apply limit to each and every function involved in the expression simultaneously.

But what you did, If I repeat but with the slight variation, you will realize that, no what you are doing is a very big mistake.

first plugin $x=0$ in the numerator's $x^{6000}$ it vanishes and you left with $\frac{sinx^{6000}}{x^2.sinx^{6000}}$

Now you left with $\lim_{x \to 0}\frac{1}{x^2}$ which is not defined.

so coming back to your way if you plugin $x=0$ to the second step you will get $(\frac{1}{0^2}-\frac{1}{0^2})$ which is not equal to zero.

there is one more example I have to explain to you and that is

calculate $lim_{x \to\infty} \frac{2}{\pi}$$(n+1)$.$\arccos\frac{1}{n}$)- $n$

one of the wrong solution similar to which you did,

plug in $x=\infty$ in $cos$ term keeping all other terms safe as it is and you will get $\frac{\pi}{2}$ canceling $\frac{2}{\pi}$ and the remaining thing is $(n+1)-(n)$ = $1$ which is the wrong answer so never plug in in some part leaving others intact, limits should be applied simultaneously. Well, I leave it as an exercise for you to find the correct answer with the correct method.

I hope this will help you. thanks

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  • $\begingroup$ check this solution, in this solution also the limits are not applied simultaneously doubtnut.com/question-answer/… $\endgroup$
    – WhyBhav
    Apr 15, 2021 at 14:53
  • $\begingroup$ It is not really necessary to apply limit to each part simultaneously. You can replace a part by its limit if the part is a term of a factor in the overall expression. Also if the part is a factor you should ensure that its limit is non-zero. $\endgroup$
    – Paramanand Singh
    Apr 15, 2021 at 15:27
  • $\begingroup$ sorry, I can't see this video, because this is not a reliable site. In this, you can easily find such videos which can completely tremble your concepts. sometimes you can get an answer by the wrong method, it is totally based on your luck. But if things are in the product and by putting the value there is not any effect in indeterminacy you can answer (always I hope this for you). I think it would be better if you ask your maths teacher or some good books. you can also see Ciapan's answer it is simple and easy to understand. once again sorry if my words are harsh. thanks $\endgroup$
    – lee
    Apr 15, 2021 at 15:28
  • $\begingroup$ @ParamanandSingh if the limit by substituting before we get non zero then it is correct then why in the example I have given got a wrong answer. $\endgroup$
    – lee
    Apr 15, 2021 at 15:31
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    $\begingroup$ what you have discussed in your detailed answer; I have written above I suppose - ( But if things are in the product and by putting the value there is not any effect in indeterminacy you can answer (always I hope this for you)) $\endgroup$
    – lee
    Apr 15, 2021 at 16:06

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