Evaluate the limit limn→∞(n+1n)n2⋅1en

Question

Evaluate the limit  limn→∞(n+1n)n2⋅1en

Addisyn Thompson · Accepted Answer

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

egowaffle26ic · Accepted Answer

Consider  limn→∞(n+1n)n2⋅1en  This limit has an indeterminate form. Let y=(1+1n)n2⋅1en. 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  limn→∞ln⁡(y)=limn→∞ln⁡(1+1n)−1n1n2  which has the indeterminate form 00. Using LHospitals Rule we see that  limn→∞ln⁡(y)=limn→∞1(1+1n)⋅−1n2+1n2−2n3  Canceling the −1n2 on the numerator and the denominator this simplifies to  limn→∞ln⁡(y)=limn→∞11+1n−12n=limn→∞−12(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)n2⋅1en=limn→∞y=limn→∞eln⁡(y)=e−12=1e  Thus

RizerMix · Accepted Answer

Setting h=1nan=(1+1n)n2=exp(ln⁡(1+1n)1n2)=exp(ln⁡(1+h)h2)Thus,limn→+∞ane−n=limh→0exp(−1h+ln⁡(1+h)h2)=e−12Given, that we know by Schwartz derivative that, if a function is C2 near x=0 we have,f″(0)2=limx→0f(x)−f(0)x−f′(0)xtaking f(x)=ln⁡(1+x), f(0)=0, f′(0)=1, f″(0)=−1−12=limh→0ln⁡(1+h)h−1h=limh→0−1h+ln⁡(1+h)h2

Evaluate the limit \lim_{n\to\infty}(\frac{n+1}{n})^{n^2}\cdot\frac{1}{e^n}

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