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->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.
Examine the following function for continuity at the indicated point.
f(x) = `{((x^2 - 9)/(x-3) "," if x ≠ 3),(6 "," if x = 3):}` at x = 3
Find the derivative of the following function from the first principle.
log(x + 1)
Find the derivative of the following function from the first principle.
ex
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
\[\lim_{x->0} \frac{e^x - 1}{x}\]=
`"d"/"dx" (1/x)` is equal to:
`"d"/"dx" ("a"^x)` =
