# f(x)=(x+1)^1/x. Calculate lim f(x); x->infinity.

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f(x)=(x+1)^1/x.

Calculate lim f(x); x->infinity.

`lim_(x->oo) (x+1)^(1/x)`

` `This is in the indeterminate form infinity raised to zero. To find the limit we need to rewrite it into a from so that we can use L'Hopital's rule.

Lets say that this limit is equal to L. Take the natural log of both sides:

ln(L) = (1/x) ln(x +1)

Focusing only on the right side we get:

`lim_(x->oo) ln(x+1)/(x)`

`<span class="AM"> </span> <br>`

` `Now we're in an infinity / infinity form and can use L'Hopital's rule.

Take the derivative of the numerator and denominator to get:

`lim_(x->oo) ((1)/(x+1))/(1)`

This limit is equal to 0.

*But we're not done yet!*

Keep in mind that the original limit we're solving for is L. This 0 is actually ln(L). To solve for L, we need to undo the logarithm by using e.

**So L = e^0 = 1. **

@jkj1362 You confused it with lim (1+x)^(1/x), when x goes to **zero**, or, equivalently, lim(1+1/x)^x, when x goes to infinity.

We see that f(x)= (x+1)^(1/x) is a decreasing function as x becomes large.f(1)= 2,f(2)=3^(1/2) =1.732 nearly and f(10)=(11)^0.1 = 1.271 and f(1001)=(1001)^0.001 may be nearly 1 if you compute.

Limit(x+1)(1/x) becomes like (inifinity)^0 , which is indetrminate form. If the limt exits , let it be L

Taking logarithms, log L = limit (1/x)log(x+1), is in ifinity/infinity form.

We use L'Hospital's rule to determine the limit, wherein we diferentiate both numerator and denominator and take limit.And if the indetermition persists, we use the technic again. We stop if there is no indetermination.

Log L = [Limit(x->inf)(log x}'/ (x)'= [Limit (1/x)] /1=1/inf=0

We conclude that log L = 0.

Therefore, L = e^0 =1.

Therefore,** Limit x->inf (x+1)^(1/x) = 1**

pavelpimen,

if f(x) = (1+x)^(1/x), limf(x) = e when x goes to infinity.

'e' is a number that represents approximately about 2.718