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My attempt:

Let $y=\left(1+\frac{1}{x}\right)^{x+3}$

Now,

$$\ln y=\ln \left(1+\frac{1}{x}\right)^{x+3}$$

$$\ln y=(x+3)\ln\left(1+\frac{1}{x}\right)$$

$$\lim_{x\to \infty} \ln y=\lim_{x\to \infty} (x+3)\ln\left(1+\frac{1}{x}\right)$$

I'm stuck. What do I do now?

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    $\begingroup$ Why not just write it as $(1+\frac{1}{x})^x(1+\frac{1}{x})^3$? $\endgroup$ Jul 14, 2022 at 11:41
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    $\begingroup$ The trick is $(x+3)ln(1+\frac{1}{x})=\frac{ln(1+\frac{1}{x})}{\frac{1}{x+3}}$ and then L'Hospital rule. $\endgroup$
    – Bob Dobbs
    Jul 14, 2022 at 11:45

5 Answers 5

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There is an easier way: $$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x+3}=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x\left(1+\frac{1}{x}\right)^3=e$$ since $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$, by definition of euler number and $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^3=1$

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Write it as $$(1+\frac{1}{x})^{x+3} = (1+\frac{1}{x})^{x}(1+\frac{1}{x})^{3}$$ Taking limit yields $$\lim_{x\to \infty}(1+\frac{1}{x})^{x}(1+\frac{1}{x})^{3} = \lim_{x\to \infty}(1+\frac{1}{x})^{x} \cdot 1 = e$$

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$$\lim_{x\to \infty} \ln y=\lim_{x\to \infty} (x+3)\ln\left(1+\frac{1}{x}\right)$$

you can write it as

$$\lim_{x\to \infty} \ln y=\lim_{x\to \infty} \frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x+3}}$$

which is of the form $$ \frac{0 }{0}$$

differentiating right side w.r.t. x gives

$$\lim_{x\to \infty} \ln y=\lim_{x\to \infty} \frac{\frac{-1}{x(x+1)}}{\frac{-1}{(x+3)^2}}=1$$

complete from here.

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HINT: You have obtained $\infty\times0$. Convert it into $\frac\infty\infty$ form or $\frac00$ and then apply L'Hopital.

If you don't want to use L'Hopital then you can use series expansion for $\log$.

$$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-...$$

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Since $\ln(1+t)\sim t$ as $t\to0$, we have $$(x+3)\ln\left(1+\frac 1x\right)\sim\frac{x+3}x\sim 1\qquad \text{as }\ x \to\infty.$$ So $\ln y\to1$ and thus $\lim_{x\to\infty} y=e$.

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