lim _{x rightarrow infty} (sqrt{x^2+5x-7}-x) | vedclass

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(C) To evaluate the limit $\lim _{x \rightarrow \infty} (\sqrt{x^2+5x-7}-x)$,we rationalize the expression: $\lim _{x \rightarrow \infty} \frac{(\sqrt

Vedclass · Accepted Answer

$\frac{5}{2}$

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(C) To evaluate the limit $\lim _{x \rightarrow \infty} (\sqrt{x^2+5x-7}-x)$,we rationalize the expression: $\lim _{x \rightarrow \infty} \frac{(\sqrt{x^2+5x-7}-x)(\sqrt{x^2+5x-7}+x)}{\sqrt{x^2+5x-7}+x}$ $= \lim _{x \rightarrow \infty} \frac{x^2+5x-7-x^2}{\sqrt{x^2+5x-7}+x}$ $= \lim _{x \rightarrow \infty} \frac{5x-7}{\sqrt{x^2+5x-7}+x}$ Dividing the numerator and denominator by $x$ (where $x > 0$): $= \lim _{x \rightarrow \infty} \frac{5-\frac{7}{x}}{\sqrt{1+\frac{5}{x}-\frac{7}{x^2}}+1}$ As $x \rightarrow \infty$,$\frac{1}{x} \rightarrow 0$ and $\frac{1}{x^2} \rightarrow 0$: $= \frac{5-0}{\sqrt{1+0-0}+1} = \frac{5}{1+1} = \frac{5}{2}$

$\lim _{x \rightarrow \infty} (\sqrt{x^2+5x-7}-x) = $

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