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sin(5*x)/(2*x^2)
Limit of the function/ sin(5*x)/(2*x^2)

Limit of the function sin(5*x)/(2*x^2)

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     /sin(5*x)\
 lim |--------|
x->0+|     2  |
     \  2*x   /
lim⁡x→0+(sin⁡(5x)2x2)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) 
Limit(sin(5*x)/((2*x^2)), x, 0)
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
lim⁡x→0+sin⁡(5x)=0\lim_{x \to 0^+} \sin{\left(5 x \right)} = 0
and limit for the denominator is
lim⁡x→0+(2x2)=0\lim_{x \to 0^+}\left(2 x^{2}\right) = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(sin⁡(5x)2x2)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right)
=
Let's transform the function under the limit a few
lim⁡x→0+(sin⁡(5x)2x2)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right)
=
lim⁡x→0+(ddxsin⁡(5x)ddx2x2)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(5 x \right)}}{\frac{d}{d x} 2 x^{2}}\right)
=
lim⁡x→0+(5cos⁡(5x)4x)\lim_{x \to 0^+}\left(\frac{5 \cos{\left(5 x \right)}}{4 x}\right)
=
lim⁡x→0+(54x)\lim_{x \to 0^+}\left(\frac{5}{4 x}\right)
=
lim⁡x→0+(54x)\lim_{x \to 0^+}\left(\frac{5}{4 x}\right)
=
∞\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
02468-8-6-4-2-1010-500500
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Rapid solution [src] 
oo
∞\infty
Expand and simplify
One‐sided limits [src] 
     /sin(5*x)\
 lim |--------|
x->0+|     2  |
     \  2*x   /
lim⁡x→0+(sin⁡(5x)2x2)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) 
oo
∞\infty 
= 377.431019234322
     /sin(5*x)\
 lim |--------|
x->0-|     2  |
     \  2*x   /
lim⁡x→0−(sin⁡(5x)2x2)\lim_{x \to 0^-}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) 
-oo
−∞-\infty 
= -377.431019234322
= -377.431019234322
Other limits x→0, -oo, +oo, 1
lim⁡x→0−(sin⁡(5x)2x2)=∞\lim_{x \to 0^-}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) = \infty
More at x→0 from the left
lim⁡x→0+(sin⁡(5x)2x2)=∞\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) = \infty
lim⁡x→∞(sin⁡(5x)2x2)=0\lim_{x \to \infty}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) = 0
More at x→oo
lim⁡x→1−(sin⁡(5x)2x2)=sin⁡(5)2\lim_{x \to 1^-}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) = \frac{\sin{\left(5 \right)}}{2}
More at x→1 from the left
lim⁡x→1+(sin⁡(5x)2x2)=sin⁡(5)2\lim_{x \to 1^+}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) = \frac{\sin{\left(5 \right)}}{2}
More at x→1 from the right
lim⁡x→−∞(sin⁡(5x)2x2)=0\lim_{x \to -\infty}\left(\frac{\sin{\left(5 x \right)}}{2 x^{2}}\right) = 0
More at x→-oo
Numerical answer [src] 
377.431019234322
377.431019234322
The graph
Limit of the function sin(5*x)/(2*x^2)
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