# Limit of Exponential Function

I am stuck on a question involving the limit of an exponential function, as follows

$$\lim_{z \to \infty} \left ( 1-\frac{4}{z+3} \right )^{z-2}$$

The following hint is given:

$$\text{Assume that} \qquad \lim_{x \to 0}\left ( \frac{\ln(1+x)}{x} \right )=1$$

My first thought was to address the behaviour of the function within the brackets: $$\lim_{z \to \infty} \left ( 1-\frac{4}{z+3} \right )=1$$ Of course $1^{z-2}$ as $z \to \infty$ is equal to one. I don't think this is correct. Wolfram Alpha informs me $$\lim_{z \to \infty} \left ( 1-\frac{4}{z+3} \right )^{z-2}=\frac{1}{e^{4}},$$ and I have not used the given hint. This is my first time solving a limit involving an exponent with a variable - I am missing something. Thank you for any help.

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If $L$ is the limit you wish to find, try computing $\ln(L)$. –  JavaMan Oct 11 '11 at 1:00
You can't let the two $z$'s in the expression to towards $\infty$ one by one. If that were valid, you could prove that $\lim_{x\to 0} x/x = 0$ because $\lim_{x\to 0} x = 0$ and $\lim_{x\to 0} 0/x = 0$. More abstractly, $$\lim_{x\to a}f(x,x) \ne \lim_{x\to a}\left(\lim_{y\to a}f(x,y)\right)$$ in general. –  Henning Makholm Oct 11 '11 at 1:25

This is what you can do. first let $z-2=n$. then your expression reduces to $\left(1-\frac{4}{n+5}\right)^n$. Now simplify to get $\left(\frac{n+1}{n+5}\right)^n$ and make another substitution: $x=1/n$. Then you get $$\left(\frac{x+1}{1+5x}\right)^{1/x}.$$ Note that at this point you take limit as $x$ goes to zero since you had limit as $n$ goes to infinity. take the natural log of this expression and bring the power down so that you make use of the hint given and find the the exponential to get the the answer.

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Thank you very much. I've reached this point: $$\lim x \to 0\left ( \frac{ln(1+x)}{x}-\frac{ln(1+5x)}{x} \right )$$ I know $$\lim x \to 0\left ( \frac{ln(1+x)}{x}\right )=1$$ But now I am stuck again with $$\lim x \to 0\left(-\frac{ln(1+5x)}{x} \right )$$ I will keep at it. –  Malthus Oct 11 '11 at 4:44
Note that this limit is $-5$ as you can check using L'hopital's rule. So you have $1-5=-4$, then take the exponential of $-4$ to get the result. –  smanoos Oct 11 '11 at 5:01
apply L'Hospital rule on the last limit –  pedja Oct 11 '11 at 5:32
Though I'm familiar with l'Hospital's rule we haven't reached it yet in our course (just starting with looking at the derivative as a function this week). This leads me to believe there is some method I am overlooking. –  Malthus Oct 11 '11 at 11:59

If you are allowed to use the fact that $$\lim_{y \to \infty}\left(1+\frac{a}{y}\right)^y=e^a,$$ then you can rewrite your expression as $$\lim_{x\to \infty}\left(\left(1-\frac{4}{x+3}\right)^{x+3}\right)^{\frac{x-2}{x+3}}.$$ As $x\to \infty$, $x+3\to \infty$, and therefore $$\lim_{x\to \infty}\left(1-\frac{4}{x+3}\right)^{x+3}=e^{-4}.$$ But $$\lim_{x\to\infty} \frac{x-2}{x+3}=1,$$ and the result follows.

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Taking logs you need to find $$\lim_{z \rightarrow \infty}(z-2)\log(1 - {4 \over z + 3})$$ You will then exponentiate the result. You can rewrite the above as $$\lim_{z \rightarrow \infty}(z-2)\bigg({-{4 \over z+ 3}}\bigg){\log(1 - {4 \over z + 3}) \over {-{4 \over z + 3}}}$$ As $z \rightarrow \infty$, ${\displaystyle {4 \over z + 3} \rightarrow 0}$, so...

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