Wednesday, 28 February 2018

Find the two limits without the use of l'Hospital's rule or series expansion.



I was asked to evaluate these two limits:



lim

\lim_{x\rightarrow0}\frac{e^{-x^2}+x^2-1}{\sin(3x^4)}



For the first one I tried to divide the numerator and denominator by x^3, but I can't get the answer unless I apply l'Hospital's rule or using a series expansion.



I also tried to use a substitution u=x^2 for the second limit, but I can't seem to relate anything between the exponential function and sine function.


Answer



As shown here we have that




  • \lim_{x\to0}\frac{\sin x-x}{x^3}=-\frac16


  • \lim_{x\to0}\frac{e^x-x-1}{x^2}=\frac12



For the second one we can use that



\frac{e^{-x^2}+x^2-1}{\sin(3x^4)}=\frac{e^{-x^2}+x^2-1}{-x^4}\cdot \frac{-3x^4}{\sin(3x^4)}\cdot\frac13


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