Standard limits giving wrong answers.

Question

As we all know, $$ \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)=1 $$ And I encountered many questions where I used this standard limit and it gave the right answer. However in the following questions I am unable to get the right answer by this.

I actually want to solve this limit. $$ \lim _{x \rightarrow 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right) $$ So the first step I took is to multiply by x^2 in the numerator and in the denominator of 1/sin^2x. Which gives us (x/sinx)^2 formed, and as I know, we can use standard limit and write 1 instead of it. $$ \lim _{x \rightarrow 0}\left(\frac{1}{x^2}-\frac{1}{x^2}\left(\frac{x}{\sin x}\right)^2\right) $$ $$ \lim _{x \rightarrow 0}\left(\frac{1}{x^2}-\frac{1}{x^2}\right)=0 $$

But this gives us the limit by further subtraction as 0 which is not correct, when checked by expansion method. Where am I wrong?

Below are some other examples of questions which I got wrong answers.

We also know, $$ \lim _{x \rightarrow 0} \frac{e^x-1}{x}=1 $$ So I used it in the following question: $$ \lim _{x \rightarrow 0}\left(\frac{\left(e^x-1-x\right)}{x^2}\right) $$ Here I divided by x^2 to both (e^x-1) and x^2 and then in (e^x-1)/x^2. I replaced e^x-1/x as 1 and then we cancel both 1/x and we get answer as 0 $$ \begin{aligned} & \lim _{x \rightarrow 0}\left(\left(\frac{e^x-1}{x}\right) \frac{1}{x}-\frac{1}{x}\right) \\ & =\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x}\right)=0 \end{aligned} $$

and it's also incorrect by expansion method. Please tell me what's wrong in my process.

Answer 1

As someone already pointed out, there seems to be a mistake in your factorization, if you multiply by $x^2$ the denominator and the numerator of $\frac{1}{\sin(x)^2}$ you will get $(\frac{x}{x\sin(x)})^2$. And then you will be looking at the limit $\frac{1}{x^2}(1-(\frac{x}{\sin(x)})^2)$. This limit is under an indeterminate form $\infty * 0$.

But more fundamentally, the reason why your approach is wrong is because you cannot deduce the second limit from the first. If you look at your Taylor expansion method, you will notice that the result for this limit comes from the third order term of $\sin(x)$. You are trying to solve a limit that needs information up to order 3 of the $\sin(x)$ around 0 by injecting the $\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$ which only gives you information up to order 1.

For the second question seems like you made a mistake, you should have $$\frac{e^x - 1 -x^2}{x^3} = \frac{1}{x^2}(\frac{e^x-1}{x} - x)$$ this gives you a $(+\infty)*1$ limit, so the limit is $+\infty$.

Answer 2

Lets detail this limit:

We have that $$\lim\limits_{x\rightarrow0}\left(\frac{1}{x^2}-\frac{1}{\sin^2(x)}\right)=\lim\limits_{x\rightarrow0}\frac{\sin^2(x)-x^2}{(x\sin(x))^2}\\=\lim\limits_{x\rightarrow0}\frac{\left(\frac{\sin(x)}{x}\right)^2-1}{\sin^2(x)}\\=\lim\limits_{x\rightarrow0}\frac{\left(\frac{\sin(x)}{x}\right)^2-1}{x^2}\cdot\frac{x^2}{\sin^2(x)}\\=\lim\limits_{x\rightarrow0}\frac{\frac{\sin(x)}{x}-1}{x}\cdot\frac{\frac{\sin(x)}{x}+1}{x}\cdot\frac{x^2}{\sin^2(x)}\\=\lim\limits_{x\rightarrow0}\frac{\sin(x)-x}{x^2}\cdot\frac{\sin(x)+x}{x^2}\cdot\left(\underbrace{\frac{x}{\sin(x)}}_{1}\right)^2$$ Here we have two indeterminate forms. We can calculate this by using inequalities or Taylor Expansion to get $\frac{-1}{3}$