The value of mathop {lim }limits_{x to 0} ,fr | vedclass

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(A) We evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{e^x - x - 1}{x^2}$.Substituting $x = 0$,we get $\frac{e^0 - 0 - 1}{0^2} = \frac{1 -

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$0.5$

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(A) We evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{e^x - x - 1}{x^2}$.Substituting $x = 0$,we get $\frac{e^0 - 0 - 1}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}$,which is an indeterminate form.Applying $L$'$H$ôpital's Rule by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(e^x - x - 1)}{\frac{d}{dx}(x^2)} = \mathop {\lim }\limits_{x 	o 0} \frac{e^x - 1}{2x}$.Substituting $x = 0$ again,we get $\frac{e^0 - 1}{2(0)} = \frac{0}{0}$,which is still an indeterminate form.Applying $L$'$H$ôpital's Rule once more:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(2x)} = \mathop {\lim }\limits_{x 	o 0} \frac{e^x}{2}$.Substituting $x = 0$,we get $\frac{e^0}{2} = \frac{1}{2} = 0.5$.

The value of $\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^x} - x - 1}}{{{x^2}}}$ is

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