Calculate the following limits using limit properties and known trigonometric limits:
$$\lim_{x\to 0}\frac{\sin^3x +\sin^2x + \sin x}{x^3+x^2+x}$$
So I know that $\lim\limits_{x\to 0}(\sin x/x)=1$ but finding difficulties here.
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- Rehearse the fundamental limits;
- divide through by $x$ and use the correct fundamental limit $$\lim_{x \to 0} \frac{\sin x}{x}.$$
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You can factor out $\sin{x}$ from the numerator and $x$ from the denominator, then apply the product law of limits to find that the total limit is 1, since the limit of the rest of the function can be directly evaluated as 1.
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Approximations are useful here $$\sin (x)=x $$ for small $x $
answered Nov 7 '16 at 15:52
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We can rewrite your limit function in the form:$$\dfrac{\sin x}{x}\left(\dfrac{\sin^2x +\sin x + 1}{x^2+x+1}\right)=\dfrac{\sin x}{x}\left(\dfrac{x^2}{x^2+x+1}\left(\dfrac{\sin^2x}{x^2}\right) +\dfrac{x}{x^2+x+1}\left(\dfrac{\sin x}{x}\right) + \dfrac{1}{x^2+x+1}\right)$$ in which you can easily apply the known result.
Here we do not need any approximations as this is only an algebraic manipulation.
