1. Why are these the answers to these limits?

I can't seem to figure out why the answers to these limits are what they are....can anyone help me understand where my thinking process is wrong:
lim (((e^-x)-2) / (e^1/x)) as x->0+

my thought process:
1. I try plugging in very small numbers that approach 0 into e^-x and see that the x in e^-x approaches 0, and that e^0 will equal 1.
2.for the denominator i try plugging some very small numbers in for x and see that e^(1/x) is e^infinity meaning e will get very big.

but the answer for this problem is 0. How?

I also struggle with this one:
lim x-> 0+ (e^-x) / (1 + e^(1/x)) = 0

I dont understand how the numerator will ever equal 0. e raised to any power will never make it equal zero....how is this the answer?

2. 1.

The numerator approaches -1.
The denominator approaches infinity.
Do you see that the ratio goes to zero ?

3. Hello, colerelm1!

I have an "eyeball" approach to this problem . . .

$\displaystyle \lim_{x\to0^+}\frac{e^{\text{-}x}-2}{e^{\frac{1}{x}}}$

Multiply by $\displaystyle \frac{e^x}{e^x}\!:\;\;\frac{1 - 2e^x}{e^x\cdot e^{\frac{1}{x}}}$

Numerator:
. . As $x\to0:\;(1- 2e^x) \;\to\; 1 - 2(1) \:=\:\text{-}1$

Denominator:
. . As $x\to0:\;\left(e^x\cdot e^{\frac{1}{x}}\right) \;\to\;(1)(\infty) \:=\:\infty$

Therefore . . .

4. Originally Posted by colerelm1

I also struggle with this one:
lim x-> 0+ (e^-x) / (1 + e^(1/x)) = 0

I dont understand how the numerator will ever equal 0. e raised to any power will never make it equal zero....how is this the answer?
No, the numerator is never zero (except as x approaches infinity).
This is very similar to the first one.

The numerator goes to 1, but the denominator goes to infinity.
The "ratio" of numerator to denominator approaches zero.

It's the same idea as considering where the following sequence is headed.....

$\frac{1}{1},\;\;\frac{1}{2},\;\;\frac{1}{4},\;\;\f rac{1}{8},\;\;\frac{1}{100},\;\;\frac{1}{100,000}, .......$