limx→0{loge(1+x)x2+x−1x} is equal to

Limit =limx→0loge(1+x)+x2−xx2=limx→0(x−12x2+13x3−…)+x2−xx2=12

limx→0{loge(1+x)x2+x−1x} is equal to

Limit =limx→0loge(1+x)+x2−xx2=limx→0(x−12x2+13x3−…)+x2−xx2=12

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$\lim_{x \to 0} {\frac{\log_{e} (1 + x)}{x^{2}} + \frac{x - 1}{x}}$ is equal to

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$\frac{1}{2}$

$- \frac{1}{2}$

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$Limit = \lim_{x \to 0} \frac{\log_{e} (1 + x) + x^{2} - x}{x^{2}} = \lim_{x \to 0} \frac{(x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \dots) + x^{2} - x}{x^{2}} = \frac{1}{2}$

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limx→0{loge(1+x)x2+x−1x} is equal to

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$\lim_{x \to 0} {\frac{\log_{e} (1 + x)}{x^{2}} + \frac{x - 1}{x}}$ is equal to