A Limit Question of 0/0 Uncertainty

Question

This questions of mine, I couldn't solve whatever I did. Please do not use L'hospital as it has not been taught to us and I don't think that kind of answer will be accepted in the exam.

$$\lim\limits_{x\to 0} \frac{3x + \sin^2x}{\sin2x - x^3}$$

Answer 1

Here is something to get you started:

To what is this limit evaluated:

$$\lim_{x\rightarrow 0} \frac{\sin x}{x}$$

You, probably, know the answer, that the above limit is $1$. Now, divide nominator and denominator by $x$. This should result in:

$$\lim_{x\rightarrow 0}\frac{3+ \frac{\sin^2 x}{x}}{\frac{\sin 2x}{x}- x^2}$$

Then , write that $\frac{\sin^2 x}{x}$ as $\frac{\sin x \cdot \sin x}{x}$ use the limit above I presented you with. At the denominator make a change of variable $u \mapsto 2x$ and make use of the limit I presented you with again.

I am pretty sure you can take it from here. :)

Answer 2

L'Hôpitals rule is not necessary here. Just divide the numerator and the denominator by $2x$ and exploit the fact that $$\lim\limits_{x\to 0}\frac{\sin x}{x}=1$$ Here are the steps $$\lim\limits_{x\to 0} \frac{3x + \sin^2x}{\sin2x - x^3}$$ $$=\lim\limits_{x\to 0} \frac{\frac{3}{2} + \frac{\sin^2x}{2x}}{\frac{\sin2x}{2x} - \frac{x^2}{2}}$$ $$=\frac{\frac{3}{2} + \frac12\lim\limits_{x\to 0} \frac{\sin^2x}{x}}{\lim\limits_{x\to 0} \frac{\sin2x}{2x} -\lim\limits_{x\to 0} \frac{x^2}{2}}$$ $$=\frac{\frac{3}{2} + \frac12\left(\lim\limits_{x\to 0} \frac{\sin x}{x}\right)\left(\lim\limits_{x\to 0} \sin x\right)}{\lim\limits_{x\to 0} \frac{\sin2x}{2x} -\lim\limits_{x\to 0} \frac{x^2}{2}}$$ $$=\frac{\frac{3}{2} + \frac12\cdot 1\cdot 0}{1-0}$$ $$=\frac32$$

Answer 3

For $x$ approaching zero replace $\sin^2 x$ by $x^2$ and $\sin2x$ by $2x$. Then $x$ will be the common factor for numerator and denominator. Cancel it out. You will obtain $3/2$ as a limit when $x$ tends to $0$.

Answer 4

What you basically do when using l'Hospitals rule is to use Taylor expansions of the denominator and nominator. Your example would be

$${3x + \sin^2x\over\sin2x - x^3} = {3x + (x + O(x^3))^2\over2x + O(x^3) - x^3}$$

Then it's quite obvious that the limit will become $3/2$. Now the trick if you don't get to use l'Hospital is to only use the knowledge and do the same thing without showing that we in fact uses l'Hopsital. What happens in this fraction is that $(x+o(x^2))^2 = O(x^2)$ and $x^3 = O(x^3)$ and "disappears" under limit.

To formalize this:

We know that $\lim_{x\to0}x^{-1}\sin x=1$ so we can write $\sin(x) = x\phi(x)$ where $\lim_{x\to0}\phi(x)=1$. That will take care of the ordos:

$${3x+\sin^2x\over\sin2x-x^3} = {3x + x^2\phi(x)\over 2x\phi(2x) - x^3}={3\over2}{1+x\phi(x)/3\over\phi(2x)-x^2/2}\to3/2$$

Then it's just a matter of combining the limits of the parts, the limit of the nominator and denominator of the last ratio is $1$.

Then you only hand in the part after "To formalize this:" and your examiner can't complain that you're using "forbidden" techniques (even though you do).