Advertisements
Advertisements
Question
Evaluate the following:
\[\lim_{x->∞} \frac{2x + 5}{x^2 + 3x + 9}\]
Advertisements
Solution
\[\lim_{x->∞} \frac{2x + 5}{x^2 + 3x + 9}\]
= \[\lim_{x->∞} \frac{x(2 + \frac{5}x)}{(1 + \frac{3}{x} + \frac{9}{x^2})}\]
[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
RELATED QUESTIONS
Evaluate the following:
`lim_(x->∞) (sum "n")/"n"^2`
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`
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
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
Evaluate: `lim_(x->1) ((2x - 3)(sqrtx - 1))/(2x^2 + x - 3)`
Verify the continuity and differentiability of f(x) = `{(1 - x if x < 1),((1 - x)(2 - x) if 1 <= x <= 2),(3 - x if x > 2):}` at x = 1 and x = 2.
\[\lim_{x->0} \frac{e^x - 1}{x}\]=
A function f(x) is continuous at x = a `lim_(x->"a")`f(x) is equal to:
`"d"/"dx" ("a"^x)` =
