Question

lim x->0 e^(2x-1)/x

Answer 1

\[\infty\]
or zero

Answer 2

okay how do you derive the answer?

Answer 3

2x-1/x -> +inf when x->0- and -inf when x->0+

Answer 4

\[\lim_{x\to 0} \frac{e^{(2x-1)}}{x}\]
Let's put x=0 and check
\[\frac{e^{-1}}{0}=\frac{finite}{0}\]
We have a finite/ 0 so
limit doesn't exist and goes to \(\infty\)

Answer 5

If a limit approaches infinity from both sides, then also it doesn't exist, and the function is not continuous at that point!!!!

Answer 6

The limit does not exist.
x->0+ results in \[\infty\]
x->0- results in \[-\infty\]

Therefore, since they are different, the limit does not exist.

Answer 7

okay I got that the limit is inf.

can you use L'Hopital rule here?

Answer 8

giving 2e^(2x-1)/1 = inf

Answer 9

Use L'Hopital's rule. Which is applicable when we take limts to infinity or zero, and returns inf/inf or 0/0.

Answer 10

You cannot use L'Hopital's Rule e^(-1) is not 0.

Answer 11

I see that now.

Answer 12

(e^(2x - 1))/x = (e^2x)/(x * e)
The expansion of e^2x = 1 + 2x +((2x)^2)/2!...............
(e^2x)/x = 1/x + 2 +...........
In the limit x ----> 0 The last series becomes infinity