Advertisements
Advertisements
प्रश्न
A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of salt water after t minutes (in grams per litre) is C(t) = `(30"t")/(200 + "t")`. What happens to the concentration as t → ∞?
Advertisements
उत्तर
Given the concentration of saltwater after t minutes is C(t) = `(30"t")/(200 + "t")`
To find the concentration of saltwater after t → ∞
`lim_("t" -> oo) "C"("t") = lim_("t" -> oo) (30"t")/(200 + "t")`
= `lim_("t" -> oo) (30"t")/("t"(200/"t" + 1))`
= `lim_("t" -> oo) (30"t")/(200/"t" + 1)`
= `30/(0 + 1)`
`lim_("t" -> oo) "C"("t")` = 30
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(x -> 3)[sqrt(2x + 6)/x]`
Evaluate the following limit :
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> -3) (3x + 2)` = – 7
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 1) (x^2 + x + 1)` = 3
Evaluate the following :
`lim_(x -> 0)[x/(|x| + x^2)]`
Evaluate the following :
`lim_(x -> 0) [(sqrt(1 - cosx))/x]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (cos x - 1)/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.0001 | 0.01 | 0.1 |
| f(x) | 0.04995 | 0.0049999 | 0.0004999 | – 0.0004999 | – 0.004999 | – 0.04995 |
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
Evaluate the following limits:
`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`
Evaluate the following limits:
`lim_(x -> oo) (1 + x - 3x^3)/(1 + x^2 +3x^3)`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/(sin 5x)`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Choose the correct alternative:
`lim_(x -> oo) sinx/x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =
The value of `lim_(x rightarrow 0) (sqrt((1 + x^2)) - sqrt(1 - x^2))/x^2` is ______.
