Question

Assume that

f(x)={e^-x if x≥0, 1+x if x<0

What is lim x→0 f(x), if the limit exist?

Do you understand this? :)

Answer 1

Honestly I forgot how to make piecewise functions in latex...so it wont look nice what I do lol..but

f(x) = { e^(-x), x >= 0
{1 + x, x < 0

Limit as x approaches 0

So...since this doesn't specify if we are going from the left or the right...and just the point where the function breaks into pieces...we need to compute 2 limits

limit as x approaches 0 of e^(-x) = ???
and
limit as x approaches 0 of 1 + x = ???

Once you have those 2 answers:

-->If they are equal...the limit is THAT answer
-->If they are different...the limit does not exist

Answer 2

Ok, great! :)

Answer 3

Here's the latex you're looking for:

`\begin{cases}`
`e^{-x} & \text{for } x\ge0 \\ %optional: [#ex], I use 2 in place of # below`
`x+1 & \text{for } x<0`
`\end{cases}`

\[\begin{cases}
e^{-x} & \text{for }x\ge0\\[2ex]
x+1 & \text{for }x <0
\end{cases}\]