So I'm trying to find the limit of the following function:

$$ \lim_{x \rightarrow 0}\frac{ x^3\sin^2(2x) }{ \cos (x^4-\frac{ \pi }{ 2 } )\tan (x^3+x)} $$

I am not allowed to use L'Hopital's rule here. So my intuition was to use the fact that:

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

After changing everything to $sin$ and $cos$, I ended up with:

$$ \lim_{x \rightarrow 0} { x^3\cos(x^3+x) }\frac{ \sin(2x) }{ \sin(x^4)\sin(x^3+x)} $$

I then multiplied the top and bottom as needed by the stuff inside the $sin$ to get:

$$ \lim_{x \rightarrow 0}\frac{ x^3\cos(x^3+x)(2x) }{ x^4(x^3+x)}\frac{ \sin(2x) }{ 2x }\frac{ x^4 }{ \sin(x^4) }\frac{ x^3+x }{ \sin(x^3+x) } $$

The last three are trivial. They are just $1$. How do I do the first limit though? The $x^4$ cancels but I still have a $x^3+x$ in the denominator which will give me undefined if I plug in $0$. Multiplying and dividing by $sin$ will do me no good either.

According to wolfram, the answer is $4$ for the entire limit but I'm struggling to see how can the first limit be equal to that since the other limits equal $1$. Any ideas?

  • $\begingroup$ $\sin (2x) \approx 2x, \tan (x^3+x) \approx x^3+x, \cos(x^4-{ \pi \over 2}) \approx x^4$. $\endgroup$ – copper.hat Oct 3 '15 at 5:44
  • $\begingroup$ No I can't use the taylor polynomial approximation either. Otherwise this would be a joke haha. $\endgroup$ – Future Math person Oct 3 '15 at 5:45
  • $\begingroup$ This is just ${\sin x \over x} \to 1$, which you are using. $\endgroup$ – copper.hat Oct 3 '15 at 5:46
  • $\begingroup$ Yes that's true but what about the first limit? If I plug in 0, I get 1/0 . $\endgroup$ – Future Math person Oct 3 '15 at 5:47
  • $\begingroup$ Well this limit is 4 according to wolfram so that's can't be true. $\endgroup$ – Future Math person Oct 3 '15 at 5:48

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