Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x->1) ((2x - 3)(sqrtx - 1))/(2x^2 + x - 3)`
Advertisements
उत्तर
`lim_(x->1) ((2x - 3)(sqrtx - 1))/(2x^2 + x - 3)`
`= lim_(x->1) ((2x - 3)(sqrtx - 1))/((2x + 3)(x - 1))` ..`- 6 {(3/2 = 2x + 3),((-2)/2 = x - 1):}`
`= lim_(x->1) ((2x - 3)(sqrtx - 1))/((2x + 3)(sqrtx - 1)(sqrtx + 1))` ...[∵ a2 - b2 = (a + b)(a - b)]
`= lim_(x->1) (2x - 3)/((2x + 3)(sqrtx + 1))`
`= (2(1) - 3)/([2(1) + 3][sqrt1 + 1])`
`= (-1)/((5)(2))`
`= (-1)/10`
APPEARS IN
संबंधित प्रश्न
Evaluate the following:
`lim_(x->0) (sqrt(1+x) - sqrt(1-x))/x`
If `lim_(x->a) (x^9 + "a"^9)/(x + "a") = lim_(x->3)` (x + 6), find the value of a.
If f(x) = `(x^7 - 128)/(x^5 - 32)`, then find `lim_(x-> 2)` f(x)
Let f(x) = `("a"x + "b")/("x + 1")`, if `lim_(x->0) f(x) = 2` and `lim_(x->∞) f(x) = 1`, then show that f(-2) = 0
Show that f(x) = |x| is continuous at x = 0.
If f(x)= `{((x - |x|)/x if x ≠ 0),(2 if x = 0):}` then show that `lim_(x->1)`f(x) does not exist.
`lim_(theta->0) (tan theta)/theta` =
\[\lim_{x->0} \frac{e^x - 1}{x}\]=
For what value of x, f(x) = `(x+2)/(x-1)` is not continuous?
If y = x and z = `1/x` then `"dy"/"dx"` =
