What is the limit as x approaches 0 of 1/x-1/((x+x^2))+ln(e^2+x) ?

The limit as x approaches 0 of 1/x-1/((x+x^2))+ln(e^2+x) is 3

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limit as x approaches 0 of 1/x-1/((x+x^2))+ln(e^2+x)

x \to 0 lim (\frac{1}{x} - \frac{1}{( x + x ^{2} )} + ln (e^{2} + x))

3

Solution steps

x \to 0 lim (\frac{1}{x} - \frac{1}{x + x ^{2}} + ln (e^{2} + x))

Expand

\frac{1}{x} - \frac{1}{x + x ^{2}} + ln (e^{2} + x) : \frac{1 + x ln ( x + e ^{2} ) + ln ( x + e ^{2} )}{x + 1}

= x \to 0 lim (\frac{1 + x ln ( x + e ^{2} ) + ln ( x + e ^{2} )}{x + 1})

Plug in the value

x = 0

= \frac{1 + 0 \cdot ln ( 0 + e ^{2} ) + ln ( 0 + e ^{2} )}{0 + 1}

Simplify

\frac{1 + 0 \cdot ln ( 0 + e ^{2} ) + ln ( 0 + e ^{2} )}{0 + 1} : 3

= 3

What is the limit as x approaches 0 of 1/x-1/((x+x^2))+ln(e^2+x) ?
The limit as x approaches 0 of 1/x-1/((x+x^2))+ln(e^2+x) is 3