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exp(x)/(1+x)
Limit of the function/ exp(x)/(1+x)

Limit of the function exp(x)/(1+x)

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You have entered [src] 
     /   x \
     |  e  |
 lim |-----|
x->oo\1 + x/
lim⁡x→∞(exx+1)\lim_{x \to \infty}\left(\frac{e^{x}}{x + 1}\right) 
Limit(exp(x)/(1 + x), x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
lim⁡x→∞ex=∞\lim_{x \to \infty} e^{x} = \infty
and limit for the denominator is
lim⁡x→∞(x+1)=∞\lim_{x \to \infty}\left(x + 1\right) = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→∞(exx+1)\lim_{x \to \infty}\left(\frac{e^{x}}{x + 1}\right)
=
lim⁡x→∞(ddxexddx(x+1))\lim_{x \to \infty}\left(\frac{\frac{d}{d x} e^{x}}{\frac{d}{d x} \left(x + 1\right)}\right)
=
lim⁡x→∞ex\lim_{x \to \infty} e^{x}
=
lim⁡x→∞ex\lim_{x \to \infty} e^{x}
=
∞\infty
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
02468-8-6-4-2-10104000-2000
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Rapid solution [src] 
oo
∞\infty
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(exx+1)=∞\lim_{x \to \infty}\left(\frac{e^{x}}{x + 1}\right) = \infty
lim⁡x→0−(exx+1)=1\lim_{x \to 0^-}\left(\frac{e^{x}}{x + 1}\right) = 1
More at x→0 from the left
lim⁡x→0+(exx+1)=1\lim_{x \to 0^+}\left(\frac{e^{x}}{x + 1}\right) = 1
More at x→0 from the right
lim⁡x→1−(exx+1)=e2\lim_{x \to 1^-}\left(\frac{e^{x}}{x + 1}\right) = \frac{e}{2}
More at x→1 from the left
lim⁡x→1+(exx+1)=e2\lim_{x \to 1^+}\left(\frac{e^{x}}{x + 1}\right) = \frac{e}{2}
More at x→1 from the right
lim⁡x→−∞(exx+1)=0\lim_{x \to -\infty}\left(\frac{e^{x}}{x + 1}\right) = 0
More at x→-oo
The graph
Limit of the function exp(x)/(1+x)
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