evaluate the limit using lhospitals rule lim(x→0) (e^x - x - 1)/(2x^2) question help: message instructor…

Explanation:

Step1: Check indeterminate form

When $x\rightarrow0$, $\lim_{x\rightarrow0}(e^{x}-x - 1)=e^{0}-0 - 1=0$ and $\lim_{x\rightarrow0}(2x^{2})=0$. So it's in $\frac{0}{0}$ form.

Step2: Apply L'Hospital's rule

Differentiate the numerator and denominator. The derivative of $e^{x}-x - 1$ is $e^{x}-1$, and the derivative of $2x^{2}$ is $4x$. So we get $\lim_{x\rightarrow0}\frac{e^{x}-1}{4x}$.

Step3: Check indeterminate form again

When $x\rightarrow0$, $\lim_{x\rightarrow0}(e^{x}-1)=e^{0}-1 = 0$ and $\lim_{x\rightarrow0}(4x)=0$. Still in $\frac{0}{0}$ form.

Step4: Apply L'Hospital's rule again

Differentiate the numerator and denominator again. The derivative of $e^{x}-1$ is $e^{x}$, and the derivative of $4x$ is $4$. So we get $\lim_{x\rightarrow0}\frac{e^{x}}{4}$.

Step5: Evaluate the limit

Substitute $x = 0$ into $\frac{e^{x}}{4}$, we have $\frac{e^{0}}{4}=\frac{1}{4}$.

Answer:

$\frac{1}{4}$