Question

lim_{x rightarrow infty}(1-2x)e ^{-2x}

Answer 1

Steps:\[\lim_{x \rightarrow \infty}(1-2x)(e^{-2x})\]\[\lim_{x \rightarrow \infty}\frac{1-2x}{e^{2x}}\]

Applying L'hopital's rule:

\[\lim_{x \rightarrow \infty} \frac{-2}{2.e^{2x}} = \lim_{x \rightarrow \infty}\frac{-2}{2.\infty} = \lim_{x \rightarrow \infty} \frac{-2}{\infty} = 0\]

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