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Evaluate: $$\lim_{x\to0^+}\frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6}$$

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There is an extra bracket in the numerator. Is that a typo or should there be a bracket on the left as well? – L. F. Feb 18 '13 at 13:33
Have you tried L'Hôpital's rule? – Stefan Hansen Feb 18 '13 at 13:33
up vote 5 down vote accepted

Exploiting the Taylor series of a function at the point $x=0$, we have

$$ \frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln(1+x)+x^4-x^6}$$

$$=\frac{(-3\sqrt{x}+x^3+(x^6-\frac{x^{12}}{3!}-\dots )))( (1+4\sqrt{x}+\dots)-1)}{4 (x+\frac{x^2}{2}\dots)+x^4-x^6}$$

$$ \sim \frac{(-3\sqrt{x})(4\sqrt{x})}{4x} = -3. $$

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Your limit can be written as

$$ \frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6} =\frac{(-3\sqrt{x}+o(\sqrt{x}))(4\sqrt{x}+o({\sqrt x}))}{4x+o(x)} = \frac{-12x +o(x)}{4x+o(x)} \to -3 $$

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$$\lim_{x\to0^+}\frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6} \\= \lim_{x\to0^+} \frac{-3\sqrt{x}+x^3+\sin(x^6)}{\sqrt x} \times \frac{e^{4\sqrt{x}}-1)}{\sqrt x} \times \frac{1}{4\frac{\ln (1+x)}{x} + x^3 - x^5 }$$

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