大一高等数学求极限1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限2 cotx -1/x当x趋于0的极限3 (x^3+x^2+x+1)^1/3-x 当x趋于∞时的极限
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![大一高等数学求极限1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限2 cotx -1/x当x趋于0的极限3 (x^3+x^2+x+1)^1/3-x 当x趋于∞时的极限](/uploads/image/z/12079579-67-9.jpg?t=%E5%A4%A7%E4%B8%80%E9%AB%98%E7%AD%89%E6%95%B0%E5%AD%A6%E6%B1%82%E6%9E%81%E9%99%901.%5B%E3%8F%91%28x-%CF%80%2F2%29%5D%2Ftan+x+%E5%BD%93x%E8%B6%8B%E4%BA%8E%CF%80%2F2%E6%97%B6%E7%9A%84%E6%9E%81%E9%99%902+cotx+-1%2Fx%E5%BD%93x%E8%B6%8B%E4%BA%8E0%E7%9A%84%E6%9E%81%E9%99%903+%EF%BC%88x%5E3%2Bx%5E2%2Bx%2B1%29%5E1%2F3-x+%E5%BD%93x%E8%B6%8B%E4%BA%8E%E2%88%9E%E6%97%B6%E7%9A%84%E6%9E%81%E9%99%90)
大一高等数学求极限1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限2 cotx -1/x当x趋于0的极限3 (x^3+x^2+x+1)^1/3-x 当x趋于∞时的极限
大一高等数学求极限
1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限
2 cotx -1/x当x趋于0的极限
3 (x^3+x^2+x+1)^1/3-x 当x趋于∞时的极限
大一高等数学求极限1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限2 cotx -1/x当x趋于0的极限3 (x^3+x^2+x+1)^1/3-x 当x趋于∞时的极限
1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限
=lim(x->π/2)1/(x-π/2)/sec²x
=lim(x->π/2)cos²x/(x-π/2)
=lim(x->π/2)2cosx(-sinx)/1
=0
2.
lim(x->0)cotx-1/x
=lim(x->0)xosx/sinx-1/x
=lim(x->0)(xcosx-sinx)/xsinx
=lim(x->0)(xcosx-sinx)/x²
=lim(x->0)(cosx-xsinx-cosx)/2x
=lim(x->0)(-xsinx)/2x
=lim(x->0)(-sinx)/2
=0
3.(x^3+x^2+x+1)^1/3-x
=lim(x->∞)e^ln(x^3+x^2+x+1)^1/3-x
=lim(x->∞)e^[ln(x^3+x^2+x+1)]/(3-x)
=e^lim(x->∞)1/(x^3+x^2+x+1) *(3x²+2x+1)/(-1)
=e^lim(x->∞)-(3x²+2x+1)/(x^3+x^2+x+1)
=e^lim(x->∞)-(6x+2)/(3x^2+2x+1)
=e^lim(x->∞)-(6)/(6x+2)
=e^0
=1