What is the limit as x approaches 1 of ln(9x)+e^x ?

The limit as x approaches 1 of ln(9x)+e^x is 2ln(3)+e

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limit as x approaches 1 of ln(9x)+e^x

x \to 1 lim (ln (9 x) + e^{x})

2 ln (3) + e

Decimal

4.91550 \dots

Solution steps

x \to 1 lim (ln (9 x) + e^{x})

Plug in the value

x = 1

= ln (9 \cdot 1) + e^{1}

Simplify

ln (9 \cdot 1) + e^{1} : 2 ln (3) + e

= 2 ln (3) + e

\int (3 x - 1)^{3} d x

\int_{0}^{\infty} x e^{- x} d x

4 x^{2} - 6 x

\int e^{- \frac{x}{a}} d x

\int_{- 1}^{0} (x^{5} - 4 x^{3} + 4 x + 4) d x

What is the limit as x approaches 1 of ln(9x)+e^x ?
The limit as x approaches 1 of ln(9x)+e^x is 2ln(3)+e