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2 $begingroup$ Show that any null sequence $(a_n)_{ninmathbb R}, a_nneq 0$ fulfils $$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}tag{1}$$ Was wondering if my approach is valid. Let's assume that $lim_{ntoinfty}a_n$ exists and that $lim_{ntoinfty}=0$ then we know that $lim_{ntoinfty} (sqrt{1+a_n} -1)$ also exists. So we can write: $$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{lim_{ntoinfty}sqrt{1+a_n}-1}{lim_{ntoinfty}a_n}=frac{1}{2}tag{2}$$ Now we know that we can rewrite this to $$lim_{ntoinfty}sqrt{1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{3}$$ We use the fact that the square root is continuous: $$lim_{ntoinfty}sqrt{1+a_n}-1=sqrt{lim_{ntoinfty}1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{4}$$ We get $$sqrt{1+0}-1=frac{1}{2}cdot 0tag{5}$$ So the equation does hold for any null sequ