$\\lim_{x\\to0}\\left(\\frac{1 + \\int_{0}^{x}e^{t^{2}}dt}{e^{x}-1}-\\frac{1}{\\sin x}\\right)=$_____.
$\\lim_{x\\to0}\\left(\\frac{1 + \\int_{0}^{x}e^{t^{2}}dt}{e^{x}-1}-\\frac{1}{\\sin x}\\right)=$_____.
Answer
Explanation:
Step1: Use equivalent - infinitesimals
When (x\to0), (e^{x}-1\sim x) and (\sin x\sim x). The original limit becomes (\lim_{x\to0}\left(\frac{1 + \int_{0}^{x}e^{t^{2}}dt}{x}-\frac{1}{x}\right)=\lim_{x\to0}\frac{\int_{0}^{x}e^{t^{2}}dt}{x})
Step2: Apply L'Hopital's rule
Since it is in the (\frac{0}{0}) form, by L'Hopital's rule (\lim_{x\to0}\frac{\int_{0}^{x}e^{t^{2}}dt}{x}=\lim_{x\to0}\frac{e^{x^{2}}}{1})
Step3: Calculate the limit
Substitute (x = 0) into (\frac{e^{x^{2}}}{1}), we get (e^{0}=1)
Answer:
(1)