+129899 I assume that the first one is :
lim x → −∞ 4e^x = 0
Note that, as x → −∞, e^x will grow smaller and smaller.....and if we let x assume some large negative value, e^x, will approach 0
Therefore,
lim x → −∞ 4e^x = 0
Likewise, I assume that the second one is :
lim x → ∞ 2e^x = ∞
Again, if we let x approach some large positive value, e^x will grow larger and larger. So, as x → ∞ , e^x will also approach ∞
Here's the graph of e^x that will give you some idea of the end behaviors of both functions :
https://www.desmos.com/calculator/ml0aj0en2z
+8262 I will try to help.
Lim x → −∞ 4e^x=0
because you can change it to look like
Lim x → ∞ 4/e^x
Then the denominator gets big so it goes to zero
But listen to the BIG BOSS first. I am trying to help, so I may be incorrect. CPhill, the class genius, will tell you the answer, and maybe correct me.
+129899 I assume that the first one is :
lim x → −∞ 4e^x = 0
Note that, as x → −∞, e^x will grow smaller and smaller.....and if we let x assume some large negative value, e^x, will approach 0
Therefore,
lim x → −∞ 4e^x = 0
Likewise, I assume that the second one is :
lim x → ∞ 2e^x = ∞
Again, if we let x approach some large positive value, e^x will grow larger and larger. So, as x → ∞ , e^x will also approach ∞
Here's the graph of e^x that will give you some idea of the end behaviors of both functions :
https://www.desmos.com/calculator/ml0aj0en2z
+209 Yes, sorry for the typo...copy and paste messed it up. Thank you both, I understand it perfectly now :)
+8262 You are very welcome! Ask your questions,if you are confused. We can help again!
(CPhill might get angry!)
Well, thanks for the +5 points!
