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3 years ago

How do you solve this limit of a function math problem?

Mathematics

1 answer:

hram777 [196]3 years ago

3 0

If you know that

$e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x$

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving $e$ .

For starters, we have

$\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}$

Let $y=\dfrac34(x+1)$ . Then as $x\to\infty$ , we also have $y\to\infty$ , and

$2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2$

So in terms of $y$ , the limit is equivalent to

$\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}$

Now use some of the properties of limits: the above is the same as

$\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}$

The first limit is trivial; $\dfrac1y\to0$ , so its value is 1. The second limit comes out to

$\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}$

To see why this is the case, replace $y=-z$ , so that $z\to-\infty$ as $y\to\infty$ , and

$\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e$

Then the limit we're talking about has a value of

$\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}$

# # #

Another way to do this without knowing the definition of $e$ as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

$\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)$

(where the notation means $\exp(x)=e^x$ , just to get everything on one line).

Recall that

$\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)$

if $f$ is continuous at $x=c$ . $\exp(x)$ is continuous everywhere, so we have

$\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)$

For the remaining limit, write

$\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}$

Now as $x\to\infty$ , both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to

$\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83$