If in fixed deposit the interest is calculated half yearly then rate of interest per annum will
be:-
(i) double (ii) three times (iii) half. (iv) none of these.
Answers
Answer:
iii)half
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Step-by-step explanation:
Answer:
\huge \pink{ \frac{1}{2} }
2
1
Step-by-step explanation:
Given limit can be written as
\begin{gathered}\displaystyle \lim_{ n \to \infty} \frac{\int_{0}^{ \frac{1}{n} } {x}^{2018x + 1} \: dx}{ \frac{1}{ {n}^{2} } } \\ \end{gathered}
n→∞
lim
n
2
1
∫
0
n
1
x
2018x+1
dx
Its quite easy to show that it's now 0/0 form of limit . So
Now use L hospital rule . To differentiate numerator we will use fundamental theorem of calculas .
\begin{gathered} \boxed{\red{{\frac{d}{dx} \int_{0} ^{y} f(t) \: dt = f(y). \frac{dy}{dx}} }}\\ \end{gathered}
dx
d
∫
0
y
f(t)dt=f(y).
dx
dy
Our limit becomes
\begin{gathered}\displaystyle \lim_{ n \to \infty} \: \frac{( \frac{1}{n} ) ^{2018. \frac{1}{n} + 1} . \frac{d}{dn}( \frac{1}{n} )}{ \frac{ - 2}{ {n}^{3} } } \\ \\ = \displaystyle \lim_{ n \to \infty} \: \frac{\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n} } \frac{1}{n} . ( \frac{ - 1}{n {}^{2} } )}{ \frac{ - 2}{ {n}^{3} } } \\ \\ = \displaystyle \lim_{ n \to \infty} \: \frac{\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n}} } {2} \\ \\ = \displaystyle \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ \log\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n}} \bigg \} \\ \\ = \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ \frac{2018}{n} \log\bigg( \frac{1}{n} \bigg) \bigg \} \\ \\ again \: l \: hopital \: rule \\ \\ = \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ 2018\bigg( \frac{1}{ \frac{1}{n} } \bigg) .\frac{ - 1}{ {n}^{2} } \bigg \} \\ \\ = \frac{1}{2} \exp \{0 \} = \frac{1}{2} \end{gathered}
n→∞
lim
n
3
−2
(
n
1
)
2018.
n
1
+1
.
dn
d
(
n
1
)
=
n→∞
lim
n
3
−2
(
n
1
)
2018.
n
1
n
1
.(
n
2
−1
)
=
n→∞
lim
2
(
n
1
)
2018.
n
1
=
2
1
n→∞
lim
exp{log(
n
1
)
2018.
n
1
}
=
2
1
n→∞
lim
exp{
n
2018
log(
n
1
)}
againlhopitalrule
=
2
1
n→∞
lim
exp{2018(
n
1
1
).
n
2
−1
}
=
2
1
exp{0}=
2
1