Advertisements
Advertisements
प्रश्न
Evaluate the following:
`lim_(x->a) (x^(5/8) - a^(5/8))/(x^(2/3) - a^(2/3))`
Advertisements
उत्तर
`lim_(x->a) (x^(5/8) - a^(5/8))/(x^(2/3) - a^(2/3))`
`["Divide both numerator and denominator by x – a"; lim_(x->a) (x^n - a^n)/(x - a) = na^n]`
= `(lim_(x->a)(x^(5/8)-a^(5/8))/(x-a))/(lim_(x->a)(x^(2/3)-a^(2/3))/(x-a))`
`= (5/8 a^(5/8 - 1))/(2/3 a^(2/3 - 1))`
`= 5/8 xx 3/2 xx ("a"^((-3)/8))/("a"^((-1)/3))`
`= 15/16 xx "a"^(-1/24)`
APPEARS IN
संबंधित प्रश्न
Evaluate the following:
`lim_(x->∞) (sum "n")/"n"^2`
If `lim_(x->a) (x^9 + "a"^9)/(x + "a") = lim_(x->3)` (x + 6), find the value of a.
If `lim_(x->2) (x^n - 2^n)/(x-2) = 448`, then find the least positive integer n.
If f(x) = `(x^7 - 128)/(x^5 - 32)`, then find `lim_(x-> 2)` f(x)
Examine the following function for continuity at the indicated point.
f(x) = `{((x^2 - 4)/(x-2) "," if x ≠ 2),(0 "," if x = 2):}` at x = 2
Find the derivative of the following function from the first principle.
x2
Evaluate: `lim_(x->1) ((2x - 3)(sqrtx - 1))/(2x^2 + x - 3)`
`lim_(theta->0) (tan theta)/theta` =
`"d"/"dx" (1/x)` is equal to:
If y = log x then y2 =
