Question

cos3x-cos7x/4x^2. Limit x tends to 0 . evaluate this?​

Answer 1

Answer:

Rahul borrowed a sum of money from a money lender for which he had to paowh he rate of 15% p.a. He paid ₹ 10,400 at the end of two years to clear the loan along with interest.. Find the sum he borrowed. 4. The interest on ₹ 500 amounts to ₹ 60 in 2 years at the same rate of interest per annum. In how

Step-by-step explanation:

rahul borrowed a sum of money from a money lender for which he had to pay interest at the rate of 15%pa he paid 10400 at the end of two years to clear the loan along with interest find the sum he borrowed

Solution

Verified by Toppr

Given,

Pratibha borrows Rs47000 form finance company to buy her first car

so the principal will be =47000

simple interest =17%

time for which money is borrowed =5 years

Simple Interest=

100

principal×time×rate

=

100

47000×5×17

=39950

The amount of money Pratibha will repay the finance company at the end of five years will be

=(47000+39950)

=Rs.86950

Answer 2

$\lim \limits _{x \: \to \: 0} \frac{ \cos3x - \cos7x }{4 {x}^{2} }$

$\sf \purple{by \: LHopital \: rule}$

$\implies \lim \limits _{x \: \to \: 0} \frac{ \frac{d}{dx}( \cos3x - \cos7x )}{ \frac{d}{dx} (4 {x}^{2} )}$

$\implies \lim \limits _{x \: \to \: 0} \frac{ - 3 \sin3x + 7 \sin7x }{ 8 {x} }$

$\sf \purple{applying \: LHopital \: rule \: again}$

$\implies \lim \limits _{x \: \to \: 0} \frac{ \frac{d}{dx}(- 3 \sin3x + 7 \sin7x )}{ \frac{d}{dx} (8 {x} )}$

$\implies \lim \limits _{x \: \to \: 0} \frac{ - 9\cos3x + 49 \cos7x }{ 8 }$

$\implies \frac{ - 9\cos3(0) + 49 \cos7(0) }{ 8 }$

$\implies \frac{ - 9\cos(0) + 49 \cos(0) }{ 8 }$

$\implies \frac{ - 9 + 49 }{ 8 }$

$\implies \frac{ 40 }{ 8 }$

$\orange{\therefore \lim \limits _{x \: \to \: 0} \frac{ \cos3x - \cos7x }{ 4 {x}^{2} } = 5}$

HOPE IT HELPS!!