lim _{x rightarrow infty}left[frac{x^2+x+3}{x | vedclass

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(D) Given,$L = \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$. This is of the form $1^\infty$. Using the formula $\lim _{x \right

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$e^2$

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(D) Given,$L = \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$. This is of the form $1^\infty$. Using the formula $\lim _{x \rightarrow \infty}[1+f(x)]^{g(x)} = e^{\lim _{x \rightarrow \infty} f(x) \cdot g(x)}$,we have: $L = \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3}{x^2-x+2}-1\right]^x$ $= \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3-x^2+x-2}{x^2-x+2}\right]^x$ $= \lim _{x \rightarrow \infty}\left[1+\frac{2x+1}{x^2-x+2}\right]^x$ $= e^{\lim _{x \rightarrow \infty} \frac{2x+1}{x^2-x+2} \cdot x}$ $= e^{\lim _{x \rightarrow \infty} \frac{2x^2+x}{x^2-x+2}}$ Dividing numerator and denominator by $x^2$: $= e^{\lim _{x \rightarrow \infty} \frac{2+1/x}{1-1/x+2/x^2}} = e^{\frac{2+0}{1-0+0}} = e^2$.

$\lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$ is equal to

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