nathi123 wrote:Let f(x)=[tex](\frac{1}{x})^{\frac{1}{lnx}} \Rightarrow lnf(x)=ln(\frac{1}{x})^{\frac{1}{lnx}} =\frac{1}{lnx} .ln\frac{1}{x}=\frac{ln\frac{1}{x}}{lnx}[/tex]
[tex]\lim_{x \to \infty} lnf(x)=\lim_{x \to \infty}\frac{ln\frac{1}{x}}{lnx}=\lim_{x \to \infty}\frac{\frac{1}{x}.(-\frac{1}{x^{2}})}{\frac{1}{x}}=\lim_{x \to \infty}(-\frac{1}{x^{2}})=0[/tex]. (Lopital)
[tex]\lim_{x \to \infty}f(x)=e^{0}=1[/tex].
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