StudentAidUnit 12: Limit and Continuity — Exercise 12 3
11th Class Mathematics · Unit 12: Limit and Continuity
1.A substance decays exponentially following the formula A(t) = A₀e^(-0.1t), where A₀ is the initial amount. Find the limit of A(t) as t → ∞
Given
Decay formula
A(t) = A0 e-0.1t
Formula
Apply limit
limt to infty A(t) = limt to infty A0 e-0.1t
= A0 limt to infty e-0.1t
As t to infty, e-0.1t to 0
= A0 × 0
Result
Limit
limt to infty A(t) = 0
2.A town's population is modeled by P(t) = 100,000/(1 + 9e^(-0.5t)). What is the long-term population as t → ∞
Given
Population model
P(t) = frac{100,000}{1 + 9e-0.5t
Formula
Apply limit t → ∞
limt to infty P(t) = limt to infty frac{100,000}{1 + 9e-0.5t
As t to infty, e-0.5t to 0
= limt to infty 100,0001 + 9(0)
= 100,0001 + 0
Result
Long-term population
limt to infty P(t) = 100,000
3.A company's weekly sales (in thousands) follow the function S(t) = 500t/(t+10). What is the limit of S(t) as t → ∞ and what does it represent?
Given
Sales function
S(t) = 500tt+10
Formula
Apply limit t → ∞
limt to infty S(t) = limt to infty 500tt+10
Divide by t
= limt to infty frac{500t}{t(1 + 10t)}
= limt to infty frac{500}{1 + 10t
As t to infty, 10t to 0
= 5001 + 0
Result
Limit
limt to infty S(t) = 500
4.(i) Signal strength S(d) at a distance d from tower is modeled as S(d) = 1000/d². What is the signal at d = 10? (ii) What happens to signal strength as d → ∞?
Given
Signal model
S(d) = 1000d2
(i) Put d = 10
S(10) = 1000(10)2
= 1000100
Result
Signal at d = 10
S(10) = 10
(ii) Apply limit d → ∞
limd to infty S(d) = limd to infty 1000d2
= 1000 limd to infty 1d2
As d to infty, 1d2 to 0
= 1000 × 0
Limit as d → ∞
limd to infty S(d) = 0
5.(i) A stock price grows according to P(t) = 50e^(0.05t). Find the limit of P(t) as t → ∞. (ii) Calculate the price after 10 years.
Given
Price model
P(t) = 50e0.05t
(i) Apply limit t → ∞
limt to infty P(t) = limt to infty 50e0.05t
= 50 limt to infty e0.05t
As t to infty, e0.05t to infty
Result
Limit
limt to infty P(t) = infty
(ii) Put t = 10
P(10) = 50e0.05(10)
P(10) = 50e0.5
= 50 × 1.6487
= 82.436 ≈ 82.44
Price after 10 years
P(10) ≈ 82.44
6.The factory's cost function is given as C(x) = {10x + 500 if x ≤ 100; 12x + 300 if x > 100}. Is the cost function continuous at x = 100?
Given
Cost function
C(x) = begin{cases} 10x + 500 & if x le 100 12x + 300 & if x > 100 end{cases}
At x = 100
C(x) = 10x + 500
C(100) = 10(100) + 500
= 1000 + 500 = 1500
Left-hand limit
limx to 100^- C(x) = limx to 100^- (10x + 500)
= 10(100) + 500 = 1500
Right-hand limit
limx to 100^+ C(x) = limx to 100^+ (12x + 300)
= 12(100) + 300 = 1500
As limx to 100^- C(x) = limx to 100^+ C(x), so
limx to 100 C(x) exists and equals 1500
As C(100) = limx to 100 C(x)
Result
Continuity at x = 100
Hence, C(x) is continuous at x = 100
7.(i) Inflation is modeled by I(t) = I₀e^(0.03t), where I₀ is the initial price index and t is the number of years. Find the inflation rate after 5 years if I₀ = 100. (ii) What is the expected price index after 10 years?
Given
Inflation model
I(t) = I0 e0.03t
(i) Put I₀ = 100, t = 5
I(5) = 100e0.03(5)
I(5) = 100e0.15
I(5) ≈ 116.18
Percentage increase
Inflation rate = I(t) - I0I0 × 100%
= 116.18 - 100100 × 100%
≈ 16.18100 × 100%
Result
Inflation rate after 5 years
Inflation rate ≈ 16.18%
(ii) Put I₀ = 100, t = 10
I(10) = 100e0.03(10)
I(10) = 100e0.3
I(10) ≈ 134.99
Price index after 10 years
I(10) ≈ 134.99
8.The cost to produce x units is: C(x) = {5x + 20 if x ≤ 10; 6x + 10 if x > 10}. Is the cost function continuous at x = 10?
Given
Cost function
C(x) = begin{cases} 5x + 20 & if x le 10 6x + 10 & if x > 10 end{cases}
At x = 10
C(x) = 5x + 20
C(10) = 5(10) + 20
= 50 + 20
C(10) = 70
Left-hand limit
limx to 10^- C(x) = limx to 10^- (5x + 20)
= 5(10) + 20 = 50 + 20 = 70
Right-hand limit
limx to 10^+ C(x) = limx to 10^+ (6x + 10)
= 6(10) + 10 = 60 + 10 = 70
As limx to 10^- C(x) = limx to 10^+ C(x), so
limx to 10 C(x) exists and equals 70
As C(10) = limx to 10 C(x)
Result
Continuity at x = 10
Hence, C(x) is continuous at x = 10