Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.5 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Question 1.
Evaluate the following limits, if necessary use L’ Hôpital’s Rule:
$$\lim _{x \rightarrow 0}$$ $$\frac { 1-cosx }{ x^2 }$$
Solution:
$$\lim _{x \rightarrow 0}$$ $$\frac { 1-cosx }{ x^2 }$$

Question 2.
$$\lim _{x \rightarrow ∞}$$ $$\frac { 2x^2-3 }{ x^2-5x+3 }$$
Solution:

Question 3.
$$\lim _{x \rightarrow ∞}$$ $$\frac { x }{ log x }$$
Solution:
$$\lim _{x \rightarrow ∞}$$ $$\frac { x }{ log x }$$ [ $$\frac { ∞ }{ ∞ }$$ indeterminate form
Applying L’ Hôpital’s Rule
$$\lim _{x \rightarrow ∞}$$ $$\frac { 1 }{ \frac{1}{x} }$$ = $$\lim _{x \rightarrow ∞}$$ x = ∞

Question 4.
$$\lim _{x \rightarrow \frac{π}{2}}$$ $$\frac { secx }{ tanx }$$
Solution:
$$\lim _{x \rightarrow \frac{π}{2}}$$ $$\frac { secx }{ tanx }$$ [ $$\frac { ∞ }{ ∞ }$$ indeterminate form
Simplifying, we get
$$\lim _{x \rightarrow \frac{π}{2}}$$ $$\frac { 1 }{ sinx }$$ = $$\frac { 1 }{ sin \frac{π}{2} }$$ = 1

Question 5.
$$\lim _{x \rightarrow ∞}$$ e-x√x
Solution:
$$\lim _{x \rightarrow ∞}$$ e-x√x [0 × ∞ indeterminate form
The other form is $$\lim _{x \rightarrow ∞}$$ $$\frac { √x }{ e^x }$$
[0 × ∞ indeterminate form
Applying L’ Hôpital’s Rule
= $$\lim _{x \rightarrow ∞}$$ $$\frac { 1 }{ 2 \sqrt{xe^x} }$$
= 0

Question 6.
$$\lim _{x \rightarrow ∞}$$ ($$\frac { 1 }{ sinx }$$ – $$\frac { 1 }{ x }$$)
Solution:

Question 7.
$$\lim _{x \rightarrow 1}$$ ($$\frac { 2 }{ x^2-1 }$$ – $$\frac { x }{ x-1 }$$)
Solution:

Question 8.
$$\lim _{x \rightarrow 0^+}$$ xx
Solution:
$$\lim _{x \rightarrow 0^+}$$ xx [0° indeterminate form
Let g(x) = xx
Taking log on both sides
log g(x) = log xx
log g(x) = x log x
$$\lim _{x \rightarrow 0^+}$$ log g(x) = $$\lim _{x \rightarrow 0^+}$$ x log x [0 × ∞ indeterminate form

Question 9.
$$\lim _{x \rightarrow ∞}$$ (1 + $$\frac { 1 }{ x }$$) x
Solution:

Applying L’ Hôpital’s Rule

Exponentiating we get, $$\lim _{x \rightarrow ∞}$$ g(x) = e1 = e

Question 10.
$$\lim _{x \rightarrow \frac{π}{2}}$$ (sin x) tan x
Solution:
$$\lim _{x \rightarrow \frac{π}{2}}$$ (sin x)tan x [1 indeterminate form]
Let g(x) = (sin x) tan x
Taking log on both sides,
log g(x) = tan x log sin x
$$\lim _{x \rightarrow \frac{π}{2}}$$ log g(x) = $$\lim _{x \rightarrow \frac{π}{2}}$$ $$\frac { log sin x }{ cot x }$$
[ $$\frac { 0 }{ 0 }$$ Indeterminate form
Applying L’ Hôpital’s Rule
= $$\lim _{x \rightarrow \frac{π}{2}}$$ ($$\frac { cotx }{ -cosec^2x }$$) = -1
exponentiating, we get
$$\lim _{x \rightarrow \frac{π}{2}}$$ g(x) = e-1 = $$\frac { 1 }{ e }$$

Question 11.
$$\lim _{x \rightarrow 0^+}$$ (cos x) $$\frac { 1 }{ x^2 }$$
Solution:
$$\lim _{x \rightarrow 0^+}$$ (cos x) $$\frac { 1 }{ x^2 }$$ [1 indeterminate form
let g(x) = (cos x)$$\frac { 1 }{ x^2 }$$
Taking log on both sides,

Question 12.
If an initial amount A0 of money is invested at an interest rate r compounded n times a year, the value of the investment after t years is A = A0(1 + $$\frac { r }{ n }$$)nt. If the interest is compounded continuously, (that is as n → ∞), show that the amount after t years is A = A0ert.
Solution:

Applying L-Hospital’s Rule

Hence Proved.