For which pair of functions f(x) and g(x) below will the lim f(x)g(x)≠0 as x->infinity
f(x) = 10x + e^(-x); g(x) = 1/5x f(x) = x^2; g(x) = e^(-4x) f(x) =(Lnx)^3; g(x) = 1/x f(x) = sqrt(x) ; g(x) = e^(-x)
is g(x)=(1/5)x or g(x)=1/(5x) ?
oh sorry it's g(x)=1/(5x)
Yes, it's fine...
\(\color{#000000 }{ \displaystyle \lim_{x\to\infty} f(x)\times g(x) }\)
What limit will this give you if you use the first pair?
limx→∞f(x)×1/(5x)?
yes, and you forgot to substitute the f(x).
\(\color{#000000 }{ \displaystyle \lim_{x\to\infty} \left[(10x + e^{-x})\times\frac{1}{5x}\right] }\)
limx→∞ 10x + e^(-x) ×1/(5x)
please expand in the limit
10x • 1/(5x) = ? e\(^{-x}\) • 1/(5x) = ?
10x • 1/(5x) = 2 e−x • 1/(5x) = e^(-x)/(5 x)
((using the rules of exponents))
yes, and e\(^{-x}\) / (5x) = 1 / (5x e\(^x\))
So our limit is \(\color{#000000 }{ \displaystyle \lim_{x\to\infty} \left[2 + \frac{1}{5xe^x}\right] }\)
And since, \(\color{#000000 }{ \displaystyle \lim_{x\to\infty} \left[2 + \frac{1}{5xe^x}\right]=\lim_{x\to\infty} \left[2 \right]+\lim_{x\to\infty} \left[ \frac{1}{5xe^x}\right] }\) you should tell me what the limit will equal.
(evaluate each limit individually and add the limits)
did we say the limit is 0?
\(\color{#000000 }{ \displaystyle\lim_{x\to\infty} \left[(10x + e^{-x})\times\frac{1}{5x}\right] = \lim_{x\to\infty} \left[2 + \frac{1}{5xe^x}\right] \\ \displaystyle =\lim_{x\to\infty} \left[2 \right]+\lim_{x\to\infty} \left[ \frac{1}{5xe^x}\right] =2+0=2}\)
oh, I may be wrong. I am sorry. :)
\(\color{#000000 }{ \displaystyle \lim_{x\to a}(b)=b }\)
A function y=b will give you an output y=b for all x, wouldn't it?
Ans same with a limit, when you take \(\color{#000000 }{ \displaystyle \lim_{x\to \infty }(2) }\)|dw:1450381494548:dw|
chris, do you want to check other limits?
((A is the correct answer - and the only one, if this is not "for all that applies" question.))
Thank you!!!!!
Join our real-time social learning platform and learn together with your friends!