Question

Compute the limit:
(a) \[L=\lim_{x \rightarrow +\infty}(1+\frac{ 6 }{ x })^{x/9}\]
and
(b)\[L=\lim_{x \rightarrow +\infty}(1+\frac{ x }{ 7 })^{\frac{ 2 }{ \ln(x) }}\]

Answer 1

i'd bomb straight in with Bernoulli's discovery - you know the e thing

so if \(\lim\limits_{n \to \infty} (1 + \dfrac{1}{n})^ n = e\)

check out that on wiki

then use some subs

Answer 2

like \(\dfrac{1}{n} = \dfrac{ 6 }{ x }\)

Answer 3

for the second one, instinct suggests working out the limit of the log of that function first

Answer 4

Still need help?

Answer 5

you got this?

Answer 6

no....

Answer 7

\[\left(1+\frac{6}{x}\right)^{\frac{x}{9}}\]is, by definition
\[\huge e^{\frac{x}{9}\log(1+\frac{6}{x})}\]

Answer 8

find \[\huge \lim_{x\to \infty}\frac{x}{9}\log(1+\frac{6}{x})\] by lhopital

Answer 9

then your answer will be \(e\) to that power

Answer 10

do you know how to find that limit?

Answer 11

I'm struggling to find the limit..

Answer 12

ok lets do it
and first off, lets ignore the 9 in the denominator, it is just a number, and we can put in in last

Answer 13

so now the question is \[\lim_{x\to \infty}x\log(1+\frac{6}{x})\]

Answer 14

the "form" is \[\infty\times 0\] is that clear or no?

Answer 15

not really..

Answer 16

ok so lets see that it is the form\[\lim_{x\to \infty}=\infty\] right?

Answer 17

damn typo, i meant \[\lim_{x\to \infty}\color{red}x=\infty\]

Answer 18

okay, I got that

Answer 19

and \[\lim_{x\to \infty}\frac{6}{x}=0\]right?

Answer 20

Yes. Oh yeah, 1/infinity =0. I remember that now

Answer 21

making \[\lim_{x\to\infty}1+\frac{6}{x}=1\]

Answer 22

and \[\log(1)=0\]so \[\lim_{x\to \infty}\log(1+\frac{6}{x})=\log(1)=0\]

Answer 23

so is it now more or less clear that the form is \[\infty\times 0\]?

Answer 24

okay i got that part now

Answer 25

ok next, what to do with that? we can't use lhopital at the moment because it is not \[\frac{0}{0}\] or \[\frac{\infty}{\infty}\]so we do a trick of algebra

Answer 26

rewrite \[x\log(1+\frac{6}{x})\]as \[\frac{\log(1+\frac{6}{x})}{\frac{1}{x}}\]

Answer 27

is that algebra clear? make sure you get it, it is a common gimmick

Answer 28

I got that. Oh, we're creating a ratio so that we could use the l'hopital rule

Answer 29

exactly, and now it is in the form \[\frac{0}{0}\]
take the derivative top and bottom separately. don't simplify too much at this step

Answer 30

you got that or no?

Answer 31

is it this?
\[(\frac{ -6 }{ x(1+\frac{ 6 }{ x }) })+\ln(1+\frac{ 6 }{ x })/(\frac{ 1 }{ x })^{2}\]

Answer 32

no

Answer 33

the derivative of \(\frac{1}{x}\)is \[-\frac{1}{x^2}\]

Answer 34

the derivative of the log of something is , by the chain rule, one over something, times the derivative of something

Answer 35

so the derivative of the numerator is \[\frac{1}{1+\frac{6}{x}}\times -\frac{6}{x^2}\]

Answer 36

the \(-x^2\) terms cancel, leaving only \[\frac{6}{1+\frac{6}{x}}\]

Answer 37

now it should be pretty clear, after you figure out that algebra, that \[\lim_{x\to \infty}\frac{6}{1+\frac{6}{x}}=6\]

Answer 38

don't forget at this step that we still have to divide by that 9 we ignored out front, so your actual limit is \[\frac{6}{9}=\frac{2}{3}\]

Answer 39

okay... Still kinda confused.. Is it possible to use log differentiation for this limit?

Answer 40

I got this...
\[1/81\ln(1+\frac{ 6 }{ x })+(\frac{ -2 }{ 3x(1+\frac{ 6 }{ x }) })\]

Answer 41

where does the + sign come from?

Answer 42

the product rule?

Answer 43

\[\frac{d}{dx}\log(f(x))=\frac{f'(x)}{f(x)}\]

Answer 44

there is no product in the numerators, it is a composition of functions \[\log(1+\frac{6}{x})\]

Answer 45

the derivative is therefore \[\frac{1}{1+\frac{6}{x}}\times \frac{d}{dx}(1+\frac{6}{x})\]
\[=\frac{1}{1+\frac{6}{x}}\times( -\frac{6}{x^2})\]

Answer 46

Can we do this?
\[\lim_{x \rightarrow +\infty}lny=\frac{ x }{ 9 }\ln(1+\frac{ 6 }{ x })\]

Answer 47

yes, that is exactly what we are doing

Answer 48

oh... okay..
i got this now...

\[\frac{ \ln(1+\frac{ 6 }{ x }) }{ 9 }-\frac{ 2 }{ 3x(1+\frac{ 6 }{ x }) }\]

Answer 49

you still have to use lhopitals rule as written above to compute that limit
it looks like you took the derivative the whole thing

Answer 50

all the work is written above, take a look and see if you can understand the steps
none were skipped

Answer 51

I am confused... :(