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(-1+e^x)/sin(x)
Limit of the function/ (-1+e^x)/sin(x)

Limit of the function (-1+e^x)/sin(x)

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You have entered [src] 
     /      x\
     |-1 + E |
 lim |-------|
x->0+\ sin(x)/
lim⁡x→0+(ex−1sin⁡(x))\lim_{x \to 0^+}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right) 
Limit((-1 + E^x)/sin(x), x, 0)
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
lim⁡x→0+(ex−1)=0\lim_{x \to 0^+}\left(e^{x} - 1\right) = 0
and limit for the denominator is
lim⁡x→0+sin⁡(x)=0\lim_{x \to 0^+} \sin{\left(x \right)} = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(ex−1sin⁡(x))\lim_{x \to 0^+}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right)
=
Let's transform the function under the limit a few
lim⁡x→0+(ex−1sin⁡(x))\lim_{x \to 0^+}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right)
=
lim⁡x→0+(ddx(ex−1)ddxsin⁡(x))\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(e^{x} - 1\right)}{\frac{d}{d x} \sin{\left(x \right)}}\right)
=
lim⁡x→0+(excos⁡(x))\lim_{x \to 0^+}\left(\frac{e^{x}}{\cos{\left(x \right)}}\right)
=
lim⁡x→0+(excos⁡(x))\lim_{x \to 0^+}\left(\frac{e^{x}}{\cos{\left(x \right)}}\right)
=
11
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
02468-8-6-4-2-1010-500000500000
Plot the graph
Rapid solution [src] 
1
11
Expand and simplify
One‐sided limits [src] 
     /      x\
     |-1 + E |
 lim |-------|
x->0+\ sin(x)/
lim⁡x→0+(ex−1sin⁡(x))\lim_{x \to 0^+}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right) 
1
11 
= 1.0
     /      x\
     |-1 + E |
 lim |-------|
x->0-\ sin(x)/
lim⁡x→0−(ex−1sin⁡(x))\lim_{x \to 0^-}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right) 
1
11 
= 1.0
= 1.0
Other limits x→0, -oo, +oo, 1
lim⁡x→0−(ex−1sin⁡(x))=1\lim_{x \to 0^-}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right) = 1
More at x→0 from the left
lim⁡x→0+(ex−1sin⁡(x))=1\lim_{x \to 0^+}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right) = 1
lim⁡x→∞(ex−1sin⁡(x))\lim_{x \to \infty}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right)
More at x→oo
lim⁡x→1−(ex−1sin⁡(x))=−1+esin⁡(1)\lim_{x \to 1^-}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right) = \frac{-1 + e}{\sin{\left(1 \right)}}
More at x→1 from the left
lim⁡x→1+(ex−1sin⁡(x))=−1+esin⁡(1)\lim_{x \to 1^+}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right) = \frac{-1 + e}{\sin{\left(1 \right)}}
More at x→1 from the right
lim⁡x→−∞(ex−1sin⁡(x))\lim_{x \to -\infty}\left(\frac{e^{x} - 1}{\sin{\left(x \right)}}\right)
More at x→-oo
Numerical answer [src] 
1.0
1.0
The graph
Limit of the function (-1+e^x)/sin(x)
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