Advertisements
Advertisements
Question
Evaluate: `lim_(x->1) ((2x - 3)(sqrtx - 1))/(2x^2 + x - 3)`
Advertisements
Solution
`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
RELATED QUESTIONS
Evaluate the following:
`lim_(x->a) (x^(5/8) - a^(5/8))/(x^(2/3) - a^(2/3))`
Evaluate the following:
`lim_(x->0) (sin^2 3x)/x^2`
If `lim_(x->a) (x^9 + "a"^9)/(x + "a") = lim_(x->3)` (x + 6), find the value of a.
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^2 - 4x if x >= 2),(x+2 if x < 2):}`, then f(0) is
\[\lim_{x->0} \frac{e^x - 1}{x}\]=
`"d"/"dx"` (5ex – 2 log x) is equal to:
If y = log x then y2 =
`"d"/"dx" ("a"^x)` =
