Answer to Question #155784 in Calculus for Ojugbele Daniel

Question

Limit x approach infinity x[ ( 1 + a/x )raise to power of 1+1/x - x raise to power of -1/x ( x+ a ) ]

Answer 1

Here we have to solve $\lim\limits_{x\to\infin}(1+\frac{a}{x})^x$

Let us assume

$$y=(1+\frac{a}{x})^x$$

Taking natural logarithm on both sides we get,

$$\Rightarrow\ln(y)=\ln[(1+\frac{a}{x})^x]\\ \Rightarrow \ln(y)=x\ln(1+\frac{a}{x})$$

Now, taking $\lim\limits_{x\to\infin}$ on both sides we have,

$$\Rightarrow \lim\limits_{x\to\infin}\ln(y)=\lim\limits_{x\to\infin}x\ln(1+\frac{a}{x})\\$$ $$\Rightarrow\lim\limits_{x\to\infin}\ln(y)=\lim\limits_{x\to\infin}\frac{ln(1+\frac{a}{x})}{\frac{1}{x}}$$

Using L'Hospital's rule to solve the RHS of the equation, as the limit is in the form of $\frac{0}{0}$ we have,

$$\Rightarrow\lim\limits_{x\to\infin}\ln(y)=\lim\limits_{x\to\infin}\frac{\frac{1}{1+\frac{a}{x}}}{\frac{-1}{x^2}}(\frac{-a}{x^2})$$

$$\Rightarrow\lim\limits_{x\to\infin}\ln(y)=a\\$$

Now, as we can interchange $\ln$ and $\lim$ we get,

$$\Rightarrow\ln\lim\limits_{x\to\infin}(y)=a\\ \Rightarrow\lim\limits_{x\to\infin}y=e^a\\ \Rightarrow\lim\limits_{x\to\infin}(1+\frac{a}{x})^x=e^a$$