If f(x) = frac{1-x+sqrt{9x^2+10x+1}}{2x},then | vedclass

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(B) Given $f(x) = \frac{1-x+\sqrt{9x^2+10x+1}}{2x}$.To find $\lim_{x \rightarrow -1^{-}} f(x)$,let $x = -1-h$,where $h \rightarrow 0^{+}$ as $x \right

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$-1$

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(B) Given $f(x) = \frac{1-x+\sqrt{9x^2+10x+1}}{2x}$.To find $\lim_{x \rightarrow -1^{-}} f(x)$,let $x = -1-h$,where $h \rightarrow 0^{+}$ as $x \rightarrow -1^{-}$.Substituting $x = -1-h$ into the expression:$f(-1-h) = \frac{1-(-1-h) + \sqrt{9(-1-h)^2 + 10(-1-h) + 1}}{2(-1-h)}$$= \frac{2+h + \sqrt{9(1+2h+h^2) - 10 - 10h + 1}}{-2(1+h)}$$= \frac{2+h + \sqrt{9+18h+9h^2 - 10 - 10h + 1}}{-2(1+h)}$$= \frac{2+h + \sqrt{9h^2 + 8h}}{-2(1+h)}$Taking the limit as $h \rightarrow 0^{+}$:$\lim_{h \rightarrow 0^{+}} \frac{2+h + \sqrt{9h^2 + 8h}}{-2(1+h)} = \frac{2+0 + \sqrt{0}}{-2(1+0)} = \frac{2}{-2} = -1$.

If $f(x) = \frac{1-x+\sqrt{9x^2+10x+1}}{2x}$,then $\lim_{x \rightarrow -1^{-}} f(x) = $

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