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Question: What is the limit of \({\left( {\dfrac{x}{{x + 1}}} \right)^x}\) as x approaches infinity?...

What is the limit of (xx+1)x{\left( {\dfrac{x}{{x + 1}}} \right)^x} as x approaches infinity?

Explanation

Solution

This is a standard format of finding the limit of an indefinite form. Here is indefinite form which is forming is 1∞{1^\infty }. One to the power infinity is known as an indeterminate form, because it is unknown. One to the power infinity is unknown because infinity itself is endless.

Complete answer:
In the above question, we have to find the limit of (xx+1)x{\left( {\dfrac{x}{{x + 1}}} \right)^x}.
We can also write it as,
⇒lim⁡x→∞(xx+1)x\Rightarrow {\lim _{x \to \infty }}{\left( {\dfrac{x}{{x + 1}}} \right)^x}
We can convert it into the format lim⁡x→∞(1+f(x))g(x)=elim⁡x→∞f(x)g(x){\lim _{x \to \infty }}{\left( {1 + f\left( x \right)} \right)^{g\left( x \right)}} = {e^{{{\lim }_{x \to \infty }}f\left( x \right)g\left( x \right)}}
It is clear that here an indefinite form is forming as 1∞{1^\infty }.
First, we have to convert our given question in the required format. But for that we have to add and subtract −1 - 1.
⇒lim⁡x→∞(1+xx+1−1)x\Rightarrow {\lim _{x \to \infty }}{\left( {1 + \dfrac{x}{{x + 1}} - 1} \right)^x}
⇒lim⁡x→∞(1+x−(x+1)x+1)x\Rightarrow {\lim _{x \to \infty }}{\left( {1 + \dfrac{{x - \left( {x + 1} \right)}}{{x + 1}}} \right)^x}
Now on comparing above equation with lim⁡x→∞(1+f(x))g(x)=elim⁡x→∞f(x)g(x){\lim _{x \to \infty }}{\left( {1 + f\left( x \right)} \right)^{g\left( x \right)}} = {e^{{{\lim }_{x \to \infty }}f\left( x \right)g\left( x \right)}}, we get
⇒elim⁡x→∞(x−(x+1)x+1)×x\Rightarrow {e^{{{\lim }_{x \to \infty }}\left( {\dfrac{{x - \left( {x + 1} \right)}}{{x + 1}}} \right) \times x}}
⇒elim⁡x→∞(−1x+1)×x\Rightarrow {e^{{{\lim }_{x \to \infty }}\left( {\dfrac{{ - 1}}{{x + 1}}} \right) \times x}}
On simplification, we get
⇒elim⁡x→∞(−xx+1)\Rightarrow {e^{{{\lim }_{x \to \infty }}\left( {\dfrac{{ - x}}{{x + 1}}} \right)}}
Now, divide numerator and denominator by x.
⇒elim⁡x→∞(−11+1x)\Rightarrow {e^{{{\lim }_{x \to \infty }}\left( {\dfrac{{ - 1}}{{1 + \dfrac{1}{x}}}} \right)}}
Now on applying limits, we get
⇒e−1\Rightarrow {e^{ - 1}}
Therefore, the value of the above integral is e−1{e^{ - 1}}.

Note:
When would we encounter a situation like 1∞{1^\infty }. You want to find the limit of the following function as it approaches infinity. For example, lim⁡x→∞=1x{\lim _{x \to \infty }} = {1^x}. When we plug infinity into this function, we see that it takes on the indeterminate form of one to the power infinity.