I need to prove the following statement
lim
x---> Infinity
3x^5-5logex/5e^x-2x^5
I have
logex/x^5 --> 0 as x---> Infinity
not sure about the other bit.
Also need to prove the following
lim
x--->o
x^2/1-e^x2 = -1
Can anyone help?
Almost impossible to make any sense of that. Please remember, if nothing else, that the Order of Operations still works.
These are very different animals:
x^2 / 1 - e^x2
and
x^2 / (1 - e^x2)
Also, what does "e^x2" mean? Maybe $\displaystyle e^{x^{2}}$? Or something else?
The first limit:
$\displaystyle \lim_{x\to \infty}\frac{3x^5-\log(e^x)}{5e^x-2x^5}=...?$
Can we assume $\displaystyle \log(e^x)$ is the natural logarithm?
(I'm not from the USA, so I don't know if they mean the natural logaritm or the logarithm with base 10, because I always use $\displaystyle \ln$ for the natural logarithm)
To prove that:
$\displaystyle \lim_{x \to \infty} \frac{3x^5-5\log(e\cdot x)}{5\cdot e^x-2x^5}=0$
My solution would be:
$\displaystyle \lim_{x\to \infty}\frac{3x^5-5\log(e\cdot x)}{5\cdot e^x-2x^5}=\lim_{x\to \infty} \frac{3-\frac{5\log(e\cdot x)}{x^5}}{\frac{5\cdot e^x}{x^5}-2}$
In general:
$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x \to a} f(x)}{\lim_{x\to a}g(x)}$
And so:
$\displaystyle \frac{3-\lim_{x\to \infty }\frac{5\log(e\cdot x)}{x^5}}}{\lim_{x\to \infty}\frac{5\cdot e^x}{x^5}-2}}$
We can calculate now the two limits by using l'Hopitals rule:
$\displaystyle \lim_{x\to \infty}\frac{5\log(e\cdot x)}{x^5}=0$
$\displaystyle \lim_{x\to \infty}\frac{5\cdot e^x}{x^5}=\infty $
So that means we get:
$\displaystyle \lim_{x \to \infty} \frac{3}{\infty}=0$