What is the limit as x approaches-1 of 2^{1+ln(x)} ?

The limit as x approaches-1 of 2^{1+ln(x)} is Does not exist

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limit as x approaches-1 of 2^{1+ln(x)}

x \to - 1 lim (2^{1 + l n (x)})

Does not exist

Solution steps

x \to - 1 lim (2^{1 + l n (x)})

x \to a - lim f (x) = x \to a + lim f (x) = L

then

x \to a lim f (x) = L

x \to - 1 - lim (2^{1 + l n (x)}) =

Does not exist

x \to - 1 + lim (2^{1 + l n (x)}) =

Does not exist

= Does not exist

x \to a lim (\frac{x + 4}{x - 3})

x \to - 5 lim (17)

x \to - \infty lim ((x)^{3})

x \to 1 lim (\frac{x}{sin ( x )})

x \to 2 lim (\frac{2 x - 1}{6 x + 3})

What is the limit as x approaches-1 of 2^{1+ln(x)} ?
The limit as x approaches-1 of 2^{1+ln(x)} is Does not exist