According to the direct substitution method, the limit of natural exponential function in $x$ minus one minus $x$ divided by square of $x$ is indeterminate as the value of $x$ approaches zero.

$\implies$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x-1-x}{x^2}}$ $\,=\,$ $\dfrac{0}{0}$

The given function is basically a rational expression and its limit is indeterminate. The two qualities direct us to use the l’hôpital’s rule to find the limit of the given function in rational form.

The expression in the numerator is a trinomial and the expression in the denominator is a monomial but they are defined in terms of $x$. So, the expressions in both numerator and denominator should be differentiated with respect to $x$ to use the l’hospital’s rule.

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}(e^x-1-x)}{\dfrac{d}{dx}{(x^2)}}}$

The terms of the expression in the numerator are connected by a minus sign. So, the derivative can be distributed to all the terms in the numerator by the subtraction rule of the differentiation.

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}(e^x)-\dfrac{d}{dx}(1)-\dfrac{d}{dx}(x)}{\dfrac{d}{dx}{(x^2)}}}$

In the numerator, the derivative of natural exponential function can be evaluated by the derivative rule of exponential function, the derivative of one can be calculated by the derivative rule of a constant and the derivative of variable $x$ can be evaluated by the derivative rule of a variable. Similarly, the derivative of $x$ square can be calculated by the power rule of derivatives.

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x-0-1}{2 \times x^{2-1}}}$

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x-1}{2 \times x^1}}$

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x-1}{2 \times x}}$

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x-1}{2x}}$

Now, evaluate the limit of the rational function by the direct substitution method.

$\,\,=\,\,\,$ $\dfrac{e^0-1}{2(0)}$

According to the zero power rule, the mathematical constant $e$ raised to the power of zero is equal to one.

$\,\,=\,\,\,$ $\dfrac{1-1}{2 \times 0}$

$\,\,=\,\,\,$ $\dfrac{0}{0}$

According to the direct substitution, it is evaluated that the limit of natural exponential function in $x$ minus one divided by two times variable $x$ is indeterminate as the $x$ tends to $0$. So, the l’hospital’s rule should be used one more time to avoid the indeterminate form.

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}(e^x-1)}{\dfrac{d}{dx}(2x)}}$

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}(e^x-1)}{\dfrac{d}{dx}(2 \times x)}}$

The number $2$ is a factor and it multiplies the variable $x$ in denominator of the rational function. The constant number $2$ can be released from the differentiation by the constant multiple derivative rule.

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}(e^x-1)}{2 \times \dfrac{d}{dx}(x)}}$

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}(e^x)-\dfrac{d}{dx}(1)}{2 \times \dfrac{d}{dx}(x)}}$

The derivatives of natural exponential function in $x$, one and variable $x$ can be evaluated by the corresponding differentiation formulas.

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x-0}{2 \times 1}}$

$\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x}{2}}$

Now, find the limit of the mathematical constant $e$ raised to the power of $x$ divided by two as the value of $x$ approaches $0$, by the direct substitution.

$\,\,=\,\,\,$ $\dfrac{e^0}{2}$

$\,\,=\,\,\,$ $\dfrac{1}{2}$

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