Since limΔx→0sinΔxΔx=1\lim\limits_{\Delta x\to 0}\frac{\sin\Delta x}{\Delta x}=1Δx→0limΔxsinΔx=1, we obtain that Δx\Delta xΔx is a first approximation of sinΔx\sin\Delta xsinΔx. Then sinΔx≈Δx\sin\Delta x\approx\Delta xsinΔx≈Δx is true.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment