Limits at Infinity...4

[nycmathguy]

Calculus
Section 2.6

 

[MathLover1]

lim(e^(1/x)) as x->0 from the left

note: e^(1/x)=

.....so, if x->0 we have

so, left limit is zero

but if x->0 from the right, limit is infinity

 

[nycmathguy]

lim(e^(1/x)) as x->0 from the left

note: e^(1/x)= View attachment 3214

View attachment 3213.....so, if x->0 we have View attachment 3215

so, left limit is zero

but if x->0 from the right, limit is infinity

View attachment 3216

The sqrt{e} when the index is 0 = e^(1/0) when replacing x with 0. I thought 1/0 = undefined. No?
The graph explains it slightly better but what you said concerning x approaching 0 is not too clear.

 

[MathLover1]

-> right limit

->left limit

since not equal, two-sided limit does not exist

 

[nycmathguy]

View attachment 3238-> right limit

View attachment 3239->left limit

since not equal, two-sided limit does not exist

If I replace x with 0 for sqrt{e}, is this not e^(1/0)?
This means e^(undefined). I understand the graph but not what you said about sqrt{e} as x tends to 0 from the left.

 

[MathLover1]

e^(1/x) is same as

(you should know that), then observe the graph to see a left and right limit

 

[nycmathguy]

e^(1/x) is same as View attachment 3245 (you should know that), then observe the graph to see a left and right limit

I know that, Mira. As x tends to 0 from the left or right, we replace x with 0, right?

So, sqrt{e} when x is the index leads to e^(1/x). Yes?
If x = 0, do we or do we not get e^(1/0)? If this is true, we get e^(undefined). No?

 

[MathLover1]

e^(1/x) is NOT sqrt{e} ........take a good look : it is

 

[nycmathguy]

e^(1/x) is NOT sqrt{e} ........take a good look : it is

The sqrt{e} = e^(1/2). No? The denominator of the exponent represents the index of the square root. No?