Advertisements
Advertisements
प्रश्न
Evaluate the following:
\[\lim_{x->∞} \frac{2x + 5}{x^2 + 3x + 9}\]
Advertisements
उत्तर
\[\lim_{x->∞} \frac{2x + 5}{x^2 + 3x + 9}\]
= \[\lim_{x->∞} \frac{x(2 + \frac{5}x)}{(1 + \frac{3}{x} + \frac{9}{x^2})}\]
[Takeout x from numerator and take x2 from the denominator]
= \[\lim_{x->∞} \frac{1}{x} \frac{(2 + \frac{5}x)}{(1 + \frac{3}{x} + \frac{9}{x^2})}\]
= `0 ((2 + 0)/(1 + 0 + 0))`
= 0
APPEARS IN
संबंधित प्रश्न
Evaluate the following:
`lim_(x->a) (x^(5/8) - a^(5/8))/(x^(2/3) - a^(2/3))`
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
Find the derivative of the following function from the first principle.
ex
Evaluate: `lim_(x->1) ((2x - 3)(sqrtx - 1))/(2x^2 + x - 3)`
Show that the function f(x) = 2x - |x| is continuous at x = 0
If f(x) = `{(x^2 - 4x if x >= 2),(x+2 if x < 2):}`, then f(0) is
For what value of x, f(x) = `(x+2)/(x-1)` is not continuous?
If y = e2x then `("d"^2"y")/"dx"^2` at x = 0 is:
