Why my method of solving this question is incorrect ? Evaluate: $\lim_{x→0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$. [closed]

Question

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.

Answer 1

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 $\lim_{x \to 0} \frac{1}{x^2}.$

Answer 2

$$\lim_{x\to a} f(x)\cdot g(x)=\lim_{x\to a} f(x)\cdot\lim_{x\to a}g(x)$$ 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 $$\lim_{x\to0}\dfrac{\sin x}x=1$$

Answer 3

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 $$\lim_{x\to 0} \frac xx \color{red}= \lim \frac 0x = \lim 0 = 0$$

Answer 4

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