can someone teach me how can I solve this limit without using the L'Hopital's Rule?
lim
Thanks in advance.
Answer
\frac{2+x^2}{2-x^2}=1+\frac{2x^2}{2-x^2}=1+\frac{x^2}{1-\frac{x^2}{2}}
and y=\frac{x^2}{1-\frac{x^2}{2}}\to 0, as x\to 0. Hence
\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{x^2}{2}}{x^2}}=(1+y)^{1/y}\to e,
as x\to 0.
Finally
\left(\frac{2+x^2}{2-x^2}\right)^{1/x^2}=\left(\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{x^2}{2}}{x^2}}\right)^{\frac{1}{1-\frac{x^2}{2}}}\to e,
since, if f(x)\to e and g(x)\to 1, then f(x)^{g(x)}\to e.
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