Evaluate the limit limn→∞(n+1n)n2⋅1en

poveli1e

poveli1e

Answered

2022-01-23

Evaluate the limit
limn(n+1n)n21en

Answer & Explanation

Addisyn Thompson

Addisyn Thompson

Expert

2022-01-24Added 16 answers

First, we get it into a form we can use LHospital
egowaffle26ic

egowaffle26ic

Expert

2022-01-25Added 7 answers

Consider
limn(n+1n)n21en
This limit has an indeterminate form. Let y=(1+1n)n21en. Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain ln(y)=n2ln(1+1n)n. We can rewrite this as ln(y)=ln(1+1n)1n1n2
The
limnln(y)=limnln(1+1n)1n1n2
which has the indeterminate form 00. Using LHospitals Rule we see that
limnln(y)=limn1(1+1n)1n2+1n22n3
Canceling the 1n2 on the numerator and the denominator this simplifies to
limnln(y)=limn11+1n12n=limn12(1+1n)=12
So far we have computed the limit of ln(y), what we really want is the limit of y. We know that y=eln(y). So
limn(n+1n)n21en=limny=limneln(y)=e12=1e
Thus
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

Setting h=1nan=(1+1n)n2=exp(ln(1+1n)1n2)=exp(ln(1+h)h2)Thus,limn+anen=limh0exp(1h+ln(1+h)h2)=e12Given, that we know by Schwartz derivative that, if a function is C2 near x=0 we have,f(0)2=limx0f(x)f(0)xf(0)xtaking f(x)=ln(1+x), f(0)=0, f(0)=1, f(0)=112=limh0ln(1+h)h1h=limh01h+ln(1+h)h2

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