# Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x-1-x}{x^2}}$ by L’Hospital’s rule

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.

### Try to find the Limit by L’Hospital’s rule

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.

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

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

### Use the L’Hôpital’s rule one more time

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.

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

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

### Evaluate the Limit by Direct substitution

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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