Question

高数题，急～！求各位帮忙！😭

Answer 1

解：
1.
(1)
( x--> 1+ ) lim { f(x) ) }= ( x--> 1+ ) lim { 1 ) = 1
( x--> 1- ) lim { f(x) ) } = ( x--> 1- ) lim { x ) = 1
( x--> -1+ ) lim { f(x) ) } = ( x--> -1+ ) lim { x ) = -1
( x--> -1- ) lim { f(x) ) } = ( x--> -1- ) lim { 1 ) = 1
(2)
( x--> 1+ ) lim { f(x) ) }= ( x--> 1- ) lim { f(x) ) = 1 ==> ( x--> 1 ) lim { f(x) ) = 1
( x--> -1+ ) lim { f(x) ) } ≠ ( x--> -1- ) lim { f(x) ) ==> ( x--> -1 ) lim { f(x) ) does not exist
2.
(3) ( x--> 2 ) lim { [x-2]/[x^2 - 3] } = [2-2]/[3^2 - 3] = 0
so ( x--> 2 ) lim { [x^2 - 3]/[x-2] } = ∞ ( limit does not exist)
(4) ( x--> 0 ) lim {[4x^3 -2x^2 + x ] / [3x^2 + x ] }
= ( x--> 0 ) lim {[4x^2 -2x +1 ] / [3x + 1 ] }
= ( x--> 0 ) lim {[4x^2 -2x +1 ] / [3x + 1 ] }
= [4* 0^2 -2*0 +1 ] / [3*0 + 1 ] = 1
(5) ( x--> ∞ ) lim { 6 - 1/x + 1/x^2} = 6 - 0 + 0 = 6
(6) ( x--> ∞ ) lim { [ x^2- 1 ] / [2x^4 -x - 1 ]}
= ( x--> ∞ ) lim { [ x^(-2)- x^(-4) ] / [2 -x^(-3) - x^(-4) ] ]}
= [ 0-0 ] / [2 - 0 - 0 ] = 0
(7)
0 ≤ / arctanx / x / ≤ (π/2) / x
but ( x--> ∞ ) lim { (π/2) / x } = 0
so ( x--> ∞ ) lim { arctanx / x } = 0
(8) ( n --> ∞ ) lim { 1 + 1/2 + 1/4 + ... +1/2^n }
= ( n --> ∞ ) lim { 2 - 1/2^n }
= 2 - 0 = 2
(9) ( n --> ∞ ) lim { 1/n^2 + 2/n^2 + ... + n/n^2 }
= ( n --> ∞ ) lim { (1/2)n(n+1)/n^2 }
= ( n --> ∞ ) lim { (1/2)( 1 + 1/n) }
= (1/2)( 1 + 0 ) } = 1/2
(10)
0 ≤ / [ (x^2 + 1) /(x^3 + x^2 ] * (2+cosx) / ≤ / [ (x^2 + 1) /(x^3 + x^2 ] / * (2+1) = 3 * / [ (x^(-1) + x^(-3)) /(1 + x^(-1) )] /
but ( x--> ∞ ) lim { 3 *[ (x^(-1) + x^(-3)) /(1 + x^(-1) } =3 * / [ (0+0) /(1+0) ] / = 0
so ( x--> ∞ ) lim { [ (x^2 + 1) /(x^3 + x^2 ] * (2+cosx) } = 0
3.
(1) ( h -->0 ) lim { [ (x+h)^2 - x^2 ] /h }
= ( h -->0 ) lim { [ 2x+h ] } = 2x
(2) ( x--> ∞ ) lim { [ (2x-1)^30 * (3x-2)^20 ] /(2x+1)^50 }
= ( x--> ∞ ) lim { [ (2-1/x)^30 * (3-2/x)^20 ] /(2+1/x)^50 }
= [ (2-0)^30 * (3-0)^20 ] /(2+0)^50 = (3/2)^20
(3)
( x--> -8 ) lim { [ (1-x)^(1/2) - 3 ] /[ 2 + x^(1/3) ] }
= ( x--> -8 ) lim { [ (1-x)^(1/2) - 3 ] * [ (1-x)^(1/2) +3 ] /( [ 2 + x^(1/3) ]* [ 2^2 - 2*x^(1/3) + x^(2/3) ] } * ( x--> -8 ) lim { [ 2^2 - 2*x^(1/3) + x^(2/3) ] /[ (1-x)^(1/2) +3 ] }
= ( x--> -8 ) lim { [ (1-x) - 3^2 ] / [ 2^2 +x ]} * ( x--> -8 ) lim { [ 2^2 - 2*x^(1/3) + x^(2/3) ] /[ (1-x)^(1/2) +3 ] }
= ( x--> -8 ) lim { - 1 } * ( x--> -8 ) lim { [ 2^2 - 2*(-8)^(1/3) + (-8)^(2/3) ] /[ (1-(-8) )^(1/2) +3 ] }
= - 1 * 2 = -2

(4)
( x -->1 ) lim { 1/(1-x) - 3/( 1-x^3 ) }
= ( x -->1 ) lim { [ x^2 + x -2 ]/ [ (1-x) ( x^2 + x +1) ] }
= ( x -->1 ) lim { [ (x+2)( x -1) ]/ [ (1-x) ( x^2 + x +1) ] }
= ( x -->1 ) lim { [ - (x+2) ]/ [ ( x^2 + x +1) ] }
= [ - (1+2) ]/ [ ( 1^2 +1 +1) ] = -1