Advertisements
Advertisements
प्रश्न
Evaluate the following limit:
`lim_(x -> 1)[(x^3 - 1)/(x^2 + 5x - 6)]`
Advertisements
उत्तर
`lim_(x -> 1)[(x^3 - 1)/(x^2 + 5x - 6)]`
= `lim_(x -> 1) ((x - 1)(x^2 + x + 1))/((x - 1)(x + 6)`
= `lim_(x -> 1)(x^2 + x + 1)/(x + 6) ...[("As" x -> 1"," x ≠ 1),(therefore x - 1 ≠ 0)]`
= `((1)^2 + 1 + 1)/(1 + 6)`
= `3/7`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limits: `lim_(x -> - 3)[(x + 3)/(x^2 + 4x + 3)]`
Evaluate the following limits: `lim_(y -> 0)[(5y^3 + 8y^2)/(3y^4 - 16y^2)]`
Evaluate the following limits: `lim_(x -> 3) [1/(x - 3) - (9x)/(x^3 - 27)]`
Evaluate the following limit:
`lim_(x -> - 2)[(x^7 + x^5 + 160)/(x^3 + 8)]`
Evaluate the following Limits: `lim_(x -> 2)[((x - 2))/(2x^2 - 7x + 6)]`
Evaluate the following Limits: `lim_(x -> 3)[(x - 3)/(sqrt(x - 2) - sqrt(4 - x))]`
Evaluate the following Limits: `lim_(x -> 4)[(3 - sqrt(5 + x))/(1 - sqrt(5 - x))]`
Evaluate the following limit :
`lim_(x -> 3) [(x^2 + 2x - 15)/(x^2 - 5x + 6)]`
Evaluate the following limit :
`lim_(x -> 3) [1/(x - 3) - (9x)/(x^3 - 27)]`
Evaluate the following limit :
`lim_(x -> 2) [(x^3 - 7x + 6)/(x^3 - 7x^2 + 16x - 12)]`
Evaluate the following limit :
`lim_(x -> 1) [(x - 2)/(x^2 - x) - 1/(x^3 - 3x^2 + 2x)]`
Evaluate the following limit:
`lim_(x -> 1) [(x^4 - 3x^2 + 2)/(x^3 - 5x^2 + 3x + 1)]`
Evaluate the following limit :
`lim_(x -> "a")[1/(x^2 - 3"a"x + 2"a"^2) + 1/(2x^2 - 3"a"x + "a"^2)]`
Evaluate the following limit:
`lim_(x-> -2) [(x^7 + x^5 + 160)/(x^3 + 8)]`
Evaluate the following limit :
`lim_(x->-2)[(x^7 + x^5 +160)/(x^3 +8)]`
Evaluate the following limit:
`lim_(z->2)[(z^2 - 5z + 6)/(z^2 - 4)]`
Evaluate the following limit:
`lim_(x->-2)[(x^7+x^5+160)/(x^3+8)]`
Evaluate the following limit:
`lim_(x-> -2)[(x^7 + x^5 + 160)/(x^3 +8)]`
