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$$\Large{\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}}$$

I know how to find this Limit only by using L’Hopital’s. Here's How:

$$={\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}}$$

$$={\lim_{x\to 0}\frac{21x^2-8x}{6x\cos(3x^2)}}$$

$$={\lim_{x\to 0}\frac{x(21x-8)}{6x\cos(3x^2)}}$$

$$={\lim_{x\to 0}\frac{(21x-8)}{6\cos(3x^2)}}$$

$$=\frac{21\cdot 0 - 8}{6 \cos 0}$$

$$=\frac{-4}{3}$$

My Question is, is there a way to do this without using the L’Hopital’s? If so, How?

Also Note: We have not been taught the series expansion of $\sin x$.

Martin Sleziak
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M.S.E
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3 Answers3

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So we have $$\lim_{x\to 0} \frac{7x^3 - 4x^2}{\sin(3x^2)}.$$ First, we have to factor the numerator so $x^3 - 4x^2 = x^2(7x - 4)$. Therefore, we have $$\lim_{x\to 0} \frac{x^2(7x - 4)}{\sin(3x^2)}.$$ After this, we multiply the numerator and the denominator by $3x^2$. Note $\frac{3x^2}{3x^2}=1$ so we aren't doing anything arbitrarily and it is still equivalent. We know that $$\lim_{x\to 0} \frac{\sin(x)}{x}=1 \text{ and } \lim_{x\to 0}\frac{x}{\sin(x)}=1$$ (you have to use a calculator or L'Hopital's theorem to check this). By doing this multiplication, we get $$\lim_{x\to 0} \frac{3x^2 x^2 (7x - 4)}{3x^2 \sin (3x^2)}.$$ And since this is similar to $x / \sin(x)$ because $x / \sin(x)$ is essentially $1 \cdot 1 / (1 \cdot \sin(1 \cdot x))$, the limit as $x$ approaches $0$ of $(x / \sin(x)) = (3x^2 / 3x^2 \sin(3x^2)$ must be true due to proportionality or similarity. After this, we can simplify the limit as $x$ approaches $0$ of $(3x^2 \cdot x^2 \cdot (7x - 4))/(3x^2\sin(3x^2))$ to the limit as $x$ approaches $0$ of $(x^2(7x - 4)/x^2)$. After this, we simplify the $x^2$ in the numerator and the denominator thus leaving the limit as $x$ approaches $0$ of $(7x -4)$ which is $-4/3$. I hope this makes sense though!

pranikc
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  • There is a really good reference for formatting the mathematics. See for example: http://math.stackexchange.com/help/notation for a very quick intro. It has links to more involved formatting, but this should get you started. – TravisJ Sep 07 '15 at 03:27
  • @coder101, I tried to format the mathematics a little more readably. If you want to see what the code looks like, just click on the "edit" button for your answer. You'll see all the code that I inserted. Most of it was just putting a $ on the beginning and end of each math statement. I also used the \frac{ }{ } to make fractions, and sine has a \ in front of it (as \sin(x)). Hope this helps. Please feel free to further clarify anything that I did not completely fix. – TravisJ Sep 07 '15 at 03:39
  • I also meant to say "I hope this makes sense though," at the end. I rushed writing this!!! Any way, I really appreciate your comment TravisJ and your update as well! I will keep it in mind :) ! – pranikc Sep 07 '15 at 21:10
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I see that this is the same thing that Travis did, but, I think, easier to read. I won't be disappointed if I don't get any upvotes.

\begin{align} \lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)} &=\lim_{x\to 0}\frac{7x-4}{3} \cdot \lim_{x\to 0}\frac{3x^2}{\sin(3x^2)}\\ &= -\frac 43 \cdot 1\\ &= -\frac 43 \end{align}

Steven Alexis Gregory
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using the fact that $\sin (3x^2) = 3x^2 + \cdots,$ you get $${\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}} = \lim_{x\to 0}\frac{-4x^2 + \cdots}{3x^2 + \cdots} = -\frac 43. $$

abel
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  • +1 Thank you for answering. However, I have not been taught the series expansion. – M.S.E Sep 07 '15 at 02:03
  • but you know that $\lim_{x \to 0} \frac{\sin x}{x} = 1,$ right? that is all $\sin x = x + \cdots$ means. – abel Sep 07 '15 at 02:05
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    Really? I dont know that :/ However I managed to solve the question :) $\lim_{x\to 0}(7x^3-4x^2)\frac{1}{\sin 3x^2}= \lim_{x\to 0} \left(\frac{7}{3}x-\frac{4}{3}\right)\frac{3x^2}{\sin 3x^2}=-\frac{4}{3}$ – M.S.E Sep 07 '15 at 02:12