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# Find the limit of $\frac{e^{5x}-1-5x}{x^2}$ as $x$ approaches 0

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$\frac{25}{2}$$\,\,\left(\approx 12.5\right) Got another answer? Verify it here ## Step-by-step Solution Problem to solve: \lim_{x\to0}\left(\frac{e^{5x}-1-5x}{x^2}\right) Choose the solving method 1 If we directly evaluate the limit \lim_{x\to 0}\left(\frac{e^{5x}-1-5x}{x^2}\right) as x tends to 0, we can see that it gives us an indeterminate form \frac{0}{0} Learn how to solve limits problems step by step online. \frac{0}{0} Learn how to solve limits problems step by step online. Find the limit of (e^(5x)-1-5x)/(x^2) as x approaches 0. If we directly evaluate the limit \lim_{x\to 0}\left(\frac{e^{5x}-1-5x}{x^2}\right) as x tends to 0, we can see that it gives us an indeterminate form. We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately. After deriving both the numerator and denominator, the limit results in. If we directly evaluate the limit \lim_{x\to 0}\left(\frac{5e^{5x}-5}{2x}\right) as x tends to 0, we can see that it gives us an indeterminate form. ## Final Answer \frac{25}{2}$$\,\,\left(\approx 12.5\right)$
SnapXam A2

### beta Got another answer? Verify it!

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e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

$\lim_{x\to0}\left(\frac{e^{5x}-1-5x}{x^2}\right)$

Limits

~ 0.1 s