evaluate the limit using lhospitals rule lim(x→0) (e^x - x - 1)/(2x^2)

Explanation:

Step1: Check indeterminate form

When (x = 0), (\lim_{x\rightarrow0}\frac{e^{x}-x - 1}{2x^{2}}=\frac{e^{0}-0 - 1}{2\times0^{2}}=\frac{1 - 1}{0}=\frac{0}{0}), so L'Hopital's rule can be applied.

Step2: Differentiate numerator and denominator

Differentiate (y_1=e^{x}-x - 1) to get (y_1'=e^{x}-1), and differentiate (y_2 = 2x^{2}) to get (y_2'=4x). Then the limit becomes (\lim_{x\rightarrow0}\frac{e^{x}-1}{4x}).

Step3: Check indeterminate form again

When (x = 0), (\lim_{x\rightarrow0}\frac{e^{x}-1}{4x}=\frac{e^{0}-1}{4\times0}=\frac{1 - 1}{0}=\frac{0}{0}), apply L'Hopital's rule again.

Step4: Differentiate numerator and denominator again

Differentiate (y_3=e^{x}-1) to get (y_3'=e^{x}), and differentiate (y_4 = 4x) to get (y_4'=4). Then the limit becomes (\lim_{x\rightarrow0}\frac{e^{x}}{4}).

Step5: Evaluate the limit

Substitute (x = 0) into (\frac{e^{x}}{4}), we get (\frac{e^{0}}{4}=\frac{1}{4}).

Answer:

(\frac{1}{4})