The value of lim _{x rightarrow infty}left(fr | vedclass

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Question

(A) The given limit is of the form $1^{\infty}$.We use the formula $\lim _{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim _{x \rightarrow a} g(x)[f(x)-1]}$.

Vedclass · Accepted Answer

$e^{2}$

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(A) The given limit is of the form $1^{\infty}$.We use the formula $\lim _{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim _{x \rightarrow a} g(x)[f(x)-1]}$.$\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x} = e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2} - 1\right)}$$= e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1 - (x^{2}-4 x+2)}{x^{2}-4 x+2}\right)}$$= e^{\lim _{x \rightarrow \infty} x \left(\frac{2 x-1}{x^{2}-4 x+2}\right)}$$= e^{\lim _{x \rightarrow \infty} \frac{2 x^{2}-x}{x^{2}-4 x+2}}$Dividing numerator and denominator by $x^{2}$,we get:$= e^{\lim _{x \rightarrow \infty} \frac{2 - 1/x}{1 - 4/x + 2/x^{2}}} = e^{2/1} = e^{2}$.

The value of $\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x}$ is

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