Advertisements
Advertisements
प्रश्न
Evaluate the following limit.
`lim_(z -> 1) (z^(1/3) - 1)/(z^(1/6) -1)`
उत्तर
We have, `lim_(z -> 1) (z^(1/3) - 1)/(z^(1/6) -1)` `(0/0 "form")`
= `lim_(z -> 1) ((z^(1/6))^2 - 1)/((z^(1/6) -1))`
= = `lim_(z -> 1) ((z^(1/6) - 1)(z^(1/6) + 1))/((z^(1/6) -1))`
= `lim_(z->1) (z^(1/6) + 1)`
= `1^(1/6) + 1`
= 1 + 1
= 2
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> 3) x + 3`
Evaluate the following limit.
`lim_(x -> pi) (x - 22/7)`
Evaluate the following limit:
`lim_(r -> l) pir^2`
Evaluate the following limit.
`lim_(x -> 4) (4x + 3)/(x - 2)`
Evaluate the following limit.
`lim_(x-> -1) (x^10 + x^5 +1)/(x -1)`
Evaluate the following limit.
`lim_(x -> 0) ((x+1)^5 - 1)/x`
Evaluate the following limit.
`lim_(x- > 2) (3x^2 - x - 10)/(x^2- 4)`
Evaluate the following limit.
`lim_(x -> 3) (x^4 - 81)/(2x^2-5x - 3)`
Evaluate the following limit.
`lim_(x-> 0) (ax + b)/(cx + 1)`
Evaluate the following limit.
`lim_(x -> 1) (ax^2 + bx + c)/(cx^2 + bx + a), a+b+c != 0`
Evaluate the following limit.
`lim_(x -> -2) (1/x + 1/2)/(x + 2)`
Evaluate the following limit.
`lim_(x -> 0) (sin ax)/ (bx)`
Evaluate the following limit.
`lim_(x -> 0) (sin ax)/(sin bx), a, b != 0`
