What is the lhopital limit as x approaches 0 of ((e^{5x}-1-5x))/((x^2)) ?

The lhopital limit as x approaches 0 of ((e^{5x}-1-5x))/((x^2)) is 25/2

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Using L'Hospital's Rule limit as x approaches 0 of (e^(5x)-1-5x)/(x^2)

lhopital

x \to 0 lim (\frac{( e ^{5 x} - 1 - 5 x )}{( x ^{2} )})

\frac{25}{2}

Solution steps

x \to 0 lim (\frac{e ^{5 x} - 1 - 5 x}{x ^{2}})

Solve using L'Hopital's Rule:

\frac{25}{2}

= \frac{25}{2}

- \frac{d}{d x} y = \frac{4}{x}

- \frac{d}{d x} (\frac{1}{x})

- \frac{d}{d x} y = 3 x

(2, 5), (3, 6)

f = 4 x + 3

What is the lhopital limit as x approaches 0 of ((e^{5x}-1-5x))/((x^2)) ?
The lhopital limit as x approaches 0 of ((e^{5x}-1-5x))/((x^2)) is 25/2