Your function is $\displaystyle e^{\frac{1}{x-1}}$ and you are asking about the limit as x goes to 1 from above and below. As x approaches 1 from above, x- 1 is positive so we have e to a larger and larger positive value. As x approaches 1 from below, x- 1 is negative so we have e to a "larger and larger" negative power. Letting y= 1/(x- 1), the limit as x approaches 1 from below is the same as $\displaystyle \lim_{y\to -\infty} e^y$. And that limit is 0.
If you want to see clearly why coming from the Left to 1 tends to zero, do this
take a number less than 1, but almost equal to 1
for example 0.9
now substitute in 1/(x - 1)
when x = 0.9, 1/(x - 1) = 1/(0.9 - 1) = 1 / -0.1 = -10
now e^(-10) = 0.000045 (which almost zero)
now take x = 0.99
after substitution
e^(-100) = 0.000000000000000000000000000000000000000000037 (wOW it is so clear as x goes to 1 from the left the function becomes zero)
try x = 0.999 if you want