anonymous one year ago 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? :)

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1. johnweldon1993

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

2. anonymous

Ok, great! :)

3. anonymous

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