Question

Find the value of k that would make the function continuous in each case.
g(x) = {(e^x-1)/x x=not0, k x=0}

Answer 1

I'm sorry this is way too complicated @Yvonmei I wish I could help I'm so srry :(

Answer 2

and, I don't see the k in any pieces of the function. Retype the question please.

Answer 3

or is it kx=0, but then you didn't specify that it is for only one x-valuer, i.e. x=0

Answer 4

I have typos, but most importantly fix the Q.

Answer 5

The function in question is this:
\[ g_1(x) = \left \{ \begin{array}{lr} e^{x}-1: & x \neq 0 \\ k: & x=0 \end{array} \right.\]
or:
\[ g_2(x) = \left \{ \begin{array}{lr} e^{x-1}: & x \neq 0 \\ k: & x=0 \end{array} \right.\]
Im going to assume it is the former (g1). In this case, for a function to be continuous, its limit must agree with the actual function value so:
\[\lim_{x \to 0} f(x) = f(0) \implies \lim_{x \to 0} e^{x}-1 = k \implies 0=k\]
So k must be zero

Answer 6

If that is the case then we get:
\[\lim_{x \to 0} f(x) = f(0) \implies \lim_{x\to 0} \frac{e^x-1}{x} = k \]
Note the limit shows an indeterminant form \( \frac{0}{0}\). Using L'hopital's rule we get:
\[\lim_{x \to 0 } \frac{e^x -1}{x} = \lim_{x \to 0 } \frac{\frac{d}{dx} \left(e^x -1 \right ) }{\frac{d}{dx} x} = \lim_{x\to 0} = \frac{e^x}{1} = 1 = k\]
So k = 1