Question

how do I evaluate this limit: lim[1+x]4/x as x->infinity

Answer 1

long divide and then evaluate the limit

Answer 2

(4x+4)/x = 4 + 4/x

so as x goes to infinity, 4 + 4/x goes to ?

Answer 3

i have no idea can you explain this to me im really bad at these

Answer 4

hint:
we can write this:
\[1 + x = 1 + 4 \cdot \left( {\frac{x}{4}} \right)\]

Answer 5

my first thought is to use
\[y=e^{\ln(y)} \]

Answer 6

sorry, the procedure is another one

Answer 7

\[e^{\ln((1+x)^\frac{4}{x}}) =e^{\frac{4}{x} \ln(1+x)}\]
now look at the limit as x goes to infinity for the exponent there

Answer 8

we can write this identity:
\[\huge {\left( {1 + x} \right)^{4/x}} = {e^{4\frac{{\ln \left( {1 + x} \right)}}{x}}}\]

Answer 9

that is @chris215 can you evaluate the following limit:
\[\lim_{x \rightarrow \infty} \frac{4 \ln(1+x)}{x}\]

Answer 10

I got 4?

Answer 11

not sure how you got that...
you know you can use l'hospital here right?

Answer 12

\[\lim_{x \rightarrow \infty}\frac{ \left( 1+x \right)4 }{x }=\lim_{x \rightarrow \infty}\frac{ 4+4x }{x }=\lim_{x \rightarrow \infty}\left[ \frac{ 4x }{x }+\frac{ 4 }{ x } \right]=\lim_{x \rightarrow \infty} \frac{ 4x }{x }+\lim_{x \rightarrow \infty} \frac{ 4 }{ x }\]\[\lim_{x \rightarrow \infty}\frac{ 4\cancel{x} }{ \cancel{x} }+\lim_{x \rightarrow \infty}\frac{ 4 }{ x }= 4+\lim_{x \rightarrow \infty}\frac{ 4 }{ x }=4+0 = 4\]

Answer 13

\[\lim_{x \rightarrow \infty} \frac{4 \ln(1+x)}{x} =\lim_{x \rightarrow \infty} \frac{\frac{d}{dx}(4 \ln(1+x))}{\frac{d}{dx}(x)}\]

Answer 14

sorry, just saw the screenshot... different problem obviously

Answer 15

\[\lim_{x \rightarrow \infty} \frac{4 \ln(1+x)}{x} =\lim_{x \rightarrow \infty} \frac{\frac{d}{dx}(4 \ln(1+x))}{\frac{d}{dx}(x)}=\lim_{x \rightarrow \infty}\frac{4 \frac{(1+x)'}{1+x}}{1}=\lim_{x \rightarrow \infty} 4 \frac{1}{1+x}=4(0) \\ =0\]
so the answer is ..
\[\lim_{x \rightarrow \infty}(1+x)^\frac{4}{x}=\lim_{x \rightarrow \infty}e^{\ln((1+x)^\frac{4}{x})} \\ =e^{\lim_{x \rightarrow \infty} \frac{4}{x} \ln(1+x)}=e^{0}=1\]

Answer 16

ok thanks im going to look back on some lessons to try and better understand this

Answer 17

the first thing to understand is
that where y>0 you can write
\[y \text{ as } e^{\ln(y)}\]

Answer 18

we usually see if this works if we have a (function of x)^(a function of x)
where these functions of x's are not constant functions

Answer 19

another example:
\[\lim_{x \rightarrow \infty}(1+\frac{2}{x})^x \\ =\lim_{x \rightarrow \infty} e^{\ln((1+\frac{2}{x})^x)} =\lim_{x \rightarrow \infty} e^{x \ln(1+\frac{2}{x})} \text{ by power rule of \log } \\ =e^{\lim_{x \rightarrow \infty} \frac{\ln(1+\frac{2}{x})}{\frac{1}{x}}} \\ \text{ now we can apply lhospital again \because we have 0/0 for that limit thingy } \\ =e^{\lim_{x \rightarrow \infty} \frac{\frac{(1+\frac{2}{x})'}{1+\frac{2}{x} } }{\frac{-1}{x^2}} }\]
\[=e^{\lim_{x \rightarrow \infty} (\frac{0+\frac{-2}{x^2}}{1+\frac{2}{x}})(-x^2)} \\ =e^{\lim_{x \rightarrow \infty} \frac{2}{1+\frac{2}{x}}} =e^{\frac{2}{1+0}}=e^{2}\]

Answer 20

actually you use the way to show:
\[\lim_{x \rightarrow \infty}(1+\frac{k}{x})^x=e^{k}\]

Answer 21

which is a famous limit I believe

Answer 22

\[\lim_{x \rightarrow \infty}\left( 1+x \right)^{\frac{ 4 }{ x }}=\lim_{x \rightarrow \infty}\left(\left( 1+x \right)^{\frac{ 1 }{ x }} \right)^4=\left(\lim_{x \rightarrow \infty}\left( 1+x \right)^{\frac{ 1 }{ x }} \right)^4\]
we only need to focus on \[\lim_{x \rightarrow \infty}\left( 1+x \right)^{\frac{ 1 }{ x }}\]
let
\[z=\left( 1+x \right)^{\frac{ 1 }{ x }} \Rightarrow \ln z= \frac{1}{x}\cdot \ln \left( 1+x\right)= \frac{\ln\left(1+x\right)}{x}\]this is an indeterminate form (infinity over infinity)
we can use l'hopittal's rule
\[\lim_{x \rightarrow \infty}\ln z = \lim_{x \rightarrow \infty}\frac{ \ln \left( 1+x \right) }{ x}= \lim_{x \rightarrow \infty}\frac{\frac{1}{ \left( 1+x \right)} }{ 1}= \lim_{x \rightarrow \infty}\frac{1 }{ x+1}=0\]
so\[\lim_{x \rightarrow \infty}\left( 1+x \right)^{\frac{ 1 }{ x }}=\lim_{x \rightarrow \infty}e^{\ln z}=e^{\lim_{x \rightarrow \infty}\ln z}=e^0 = 1\]thus\[\lim_{x \rightarrow \infty}\left( 1+x \right)^{\frac{ 4 }{ x }}=1\]

Answer 23

check this site
http://tutorial.math.lamar.edu/Problems/CalcI/LHospitalsRule.aspx