evaluate the limit using lhôpitals rule lim x→0 (e^x + 3x - 1) / 8x
evaluate the limit using lhôpitals rule lim x→0 (e^x + 3x - 1) / 8x
Answer
Explanation:
Step1: Check L'Hopital - rule condition
When (x\rightarrow0), (\lim_{x\rightarrow0}(e^{x}+3x - 1)=e^{0}+3\times0 - 1=0) and (\lim_{x\rightarrow0}(8x)=0). So, we can apply L'Hopital's rule.
Step2: Differentiate numerator and denominator
The derivative of (y = e^{x}+3x - 1) is (y^\prime=e^{x}+3), and the derivative of (y = 8x) is (y^\prime = 8). So, (\lim_{x\rightarrow0}\frac{e^{x}+3x - 1}{8x}=\lim_{x\rightarrow0}\frac{e^{x}+3}{8}).
Step3: Evaluate the new - limit
Substitute (x = 0) into (\frac{e^{x}+3}{8}), we get (\frac{e^{0}+3}{8}=\frac{1 + 3}{8}).
Answer:
(\frac{1}{2})