Question

Can anyone help me please ...!!!
induction proof :
Prove that 3^n >= 1+2n

Answer 1

Can you help with that please :
lim (5^n - 4^n)/(5^n + 4^n) as x approaches to 0 @Michele_Laino

Answer 2

what if n approaches to infinit ?

Answer 3

as n approaches to zero, we have:
\[\frac{{{5^n} - {4^n}}}{{{5^n} + {4^n}}} \to \frac{{1 - 1}}{{1 + 1}} = 0\]

Answer 4

if it approaches to infinit ? @Michele_Laino

Answer 5

if n approaches to infinity, we have:
\[\frac{{{5^n} - {4^n}}}{{{5^n} + {4^n}}} = \frac{{{5^n}}}{{{5^n}}}\frac{{1 - {{\left( {4/5} \right)}^n}}}{{1 + {{\left( {4/5} \right)}^n}}} = \frac{{1 - {{\left( {4/5} \right)}^n}}}{{1 + {{\left( {4/5} \right)}^n}}} \to \frac{{1 - 0}}{{1 + 0}} = 1\]

Answer 6

thank you so much !!! omg ... i got so many questions :((((

Answer 7

thank you! :)

Answer 8

what about this:
lim \[\sqrt{n}\div \sqrt{n+1} as n approach \to infinity\] @Michele_Laino

Answer 9

we can write:
\[\frac{{\sqrt n }}{{\sqrt {n + 1} }} = \sqrt {\frac{n}{{n + 1}}} = \sqrt {\frac{1}{{1 + \left( {1/n} \right)}}} \to \sqrt {\frac{1}{{1 + 0}}} = ...?\]

Answer 10

\[\lim [2-(1/2)^n]/[3+(e/\pi)^n]\] as n approaches to infinity

Answer 11

@Michele_Laino

Answer 12

here we have:
e< pi
since e=2.71828..
and
pi=3.14159
so we can write:
\[\frac{{2 - {{\left( {1/2} \right)}^n}}}{{3 + {{\left( {e/\pi } \right)}^n}}} \to \frac{{2 - 0}}{{3 + 0}} = ...?\]

Answer 13

hmmm ... i thought about it but it seemed so easy to be that short :)

Answer 14

induction proof :
Prove that 3^n >= 1+2n @Michele_Laino

Answer 15

if we set n=0, we have:
left side 3^0=1
right side 2*0+1=1
so 1>=1 it is true, then our relation is checked for n=0

Answer 16

now I set n=1
left side = 3^1=3
right side 1+ 2*1= 3
so 3 >=3
and our relationship is true for n=1

Answer 17

now, let's suppose that our relationship is true for a generic n. Then we evaluate our relationship for n+1. I can write this:
\[\begin{gathered}
{3^{n + 1}} = 3 \cdot {3^n} \geqslant 3\left( {1 + 2n} \right) = 3 + 6n \geqslant 3 + 2n = \hfill \\
= 1 + 2n + 2 = 1 + 2\left( {n + 1} \right) \hfill \\
\end{gathered} \]

Answer 18

so we get:
\[{3^{n + 1}} \geqslant 1 + 2\left( {n + 1} \right)\]
then our relationship is true for all natural number n

Answer 19

lim \[\sqrt{(x^2+3x+2)}/(2x+5) \] as x approaches + infinity @Michele_Laino

Answer 20

can you help me with that or no ? @Michele_Laino

Answer 21

as x goes to +infinity, your function can be rewritten as below:
\[\begin{gathered}
\frac{{\sqrt {{x^2} + 3x + 2} }}{{2x + 5}} = \frac{{x\sqrt {1 + \left( {3/x} \right) + \left( {2/{x^2}} \right)} }}{{x\left( {2 + \left( {5/x} \right)} \right)}} = \hfill \\
\hfill \\
= \frac{{\sqrt {1 + \left( {3/x} \right) + \left( {2/{x^2}} \right)} }}{{\left( {2 + \left( {5/x} \right)} \right)}} \to \frac{{\sqrt {1 + 0 + 0} }}{{2 + 0}} = ...? \hfill \\
\end{gathered} \]

Answer 22

\[\Large \begin{gathered}
\frac{{\sqrt {{x^2} + 3x + 2} }}{{2x + 5}} = \frac{{x\sqrt {1 + \left( {3/x} \right) + \left( {2/{x^2}} \right)} }}{{x\left( {2 + \left( {5/x} \right)} \right)}} = \hfill \\
\hfill \\
= \frac{{\sqrt {1 + \left( {3/x} \right) + \left( {2/{x^2}} \right)} }}{{\left( {2 + \left( {5/x} \right)} \right)}} \to \frac{{\sqrt {1 + 0 + 0} }}{{2 + 0}} = ...? \hfill \\
\end{gathered} \]

Answer 23

Thank you so so much !!! :)

Answer 24

Does the graph of function \[x ^{n}/(x ^{3}-4x+5)\] has asymptote horisontale for: a). n=3 b).n<3

Answer 25

@Michele_Laino

Answer 26

Does the graph of function \[x ^{n}/(x ^{3}-4x+5)\] has asymptote horisontale for: a). n=3 b).n<3 @Abhisar can u help me please ?

Answer 27

Does the graph of function xn/(x3−4x+5) has asymptote horisontale for: a). n=3 b).n<3
can u help me please ? @amistre64 @Math&ReadingHelp

Answer 28

sorry dont know this kind of math :(

Answer 29

please post new questions .. as a new question.