Limit of \( \displaystyle \left(2 x - 1\right)^{2} e^{1 - 2 x} \) as \( x \to \infty \)

Problem 1.122 · medium

1. \[ \lim_{x \to \infty}\left(\left(2 x - 1\right)^{2} e^{1 - 2 x}\right) \]
   limit rewriteStart with the original limit. Rewrite the exponential term using the property e^(-a) = 1/e^a.✓ Proved
2. \[ = \lim_{x \to \infty}\left(\left(4 x^{2} - 4 x + 1\right) e^{1 - 2 x}\right) \]
   algebraExpand the squared binomial.✓ Proved
3. \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(4 x^{2} - 4 x + 1\right)}{\frac{d}{d x} e^{2 x - 1}}\right) \]
   lhopitalApply L'Hopital's rule because the limit is of the form infinity/infinity.✓ Proved
4. \[ = \lim_{x \to \infty}\left(\frac{\left(8 x - 4\right) e^{1 - 2 x}}{2}\right) \]
   simplifyCompute the derivatives.✓ Proved
5. \[ = \lim_{x \to \infty}\left(\left(4 x - 2\right) e^{1 - 2 x}\right) \]
   algebraDivide the numerator and denominator by 2.✓ Proved
6. \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(4 x - 2\right)}{\frac{d}{d x} e^{2 x - 1}}\right) \]
   lhopitalApply L'Hopital's rule again.✓ Proved
7. \[ = \lim_{x \to \infty}\left(2 e^{1 - 2 x}\right) \]
   simplify algebra rewrite algebraCompute the derivatives. Simplify the fraction. Move the exponential to the numerator. Distribute the negative sign in the exponent.✓ Proved
8. \[ = 0 \]
   limitThe limit of the exponential term as x approaches infinity is 0.✓ Proved