大师作文网一道关于求极限的题目,希望给出完整解题步骤
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/11 15:07:26
一道关于求极限的题目,希望给出完整解题步骤
g.e.= lim(t→0+){[(1/t)²-1/t](e^t)-√[(1/t)^4+(1/t)²]} (令 t=1/x)
= lim(t→0+)[(1-t)(e^t)-√(1+t²)]/t²
= lim(t→0+){{(1-t)²[e^(2t)]-(1+t²)}/t²}*lim(t→0+){1/[(1-t)(e^t)+√(1+t²)]}
= (1/2)*lim(t→0+){(1-t)²[e^(2t)]-(1+t²)}/t²
= (1/2)*{lim(t→0+){[e^(2t)]-1-2t[e^(2t)]}/t² + lim(t→0+){[e^(2t)]-1}}
= (1/2)*lim(t→0+){[e^(2t)]-1-2t[e^(2t)]}/t² + 0 (0/0)
= (1/2)*lim(t→0+){2[e^(2t)]-2[e^(2t)]-2t[e^(2t)]*2}/(2t)
= (1/2)*lim(t→0+){-4[e^(2t)]}/2
= -1.
= lim(t→0+)[(1-t)(e^t)-√(1+t²)]/t²
= lim(t→0+){{(1-t)²[e^(2t)]-(1+t²)}/t²}*lim(t→0+){1/[(1-t)(e^t)+√(1+t²)]}
= (1/2)*lim(t→0+){(1-t)²[e^(2t)]-(1+t²)}/t²
= (1/2)*{lim(t→0+){[e^(2t)]-1-2t[e^(2t)]}/t² + lim(t→0+){[e^(2t)]-1}}
= (1/2)*lim(t→0+){[e^(2t)]-1-2t[e^(2t)]}/t² + 0 (0/0)
= (1/2)*lim(t→0+){2[e^(2t)]-2[e^(2t)]-2t[e^(2t)]*2}/(2t)
= (1/2)*lim(t→0+){-4[e^(2t)]}/2
= -1.