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tan(9*x)/(8*x)
Limit of the function/ tan(9*x)/(8*x)

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tan(9*x)/(8*x)

What you mean?

tan(9*x)/((8*x))
 
tan(9*x)*x/8
 
tan(9*x)/((8*x))
 
tan(9*x)*x/8
 
tan(9*x)/((8*x))
 
tan(9*x)*x/8
 
tan(9*x)*x/8

Limit of the function tan(9*x)/(8*x)

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The solution

You have entered [src] 
     /tan(9*x)\
 lim |--------|
x->0+\  8*x   /
lim⁡x→0+(tan⁡(9x)8x)\lim_{x \to 0^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) 
Limit(tan(9*x)/((8*x)), x, 0)
Detail solution
Let's take the limit
lim⁡x→0+(tan⁡(9x)8x)\lim_{x \to 0^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right)
transform
lim⁡x→0+(tan⁡(9x)8x)=lim⁡x→0+(18xsin⁡(9x)cos⁡(9x))\lim_{x \to 0^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) = \lim_{x \to 0^+}\left(\frac{\frac{1}{8 x} \sin{\left(9 x \right)}}{\cos{\left(9 x \right)}}\right)
=
lim⁡x→0+(18xsin⁡(9x))lim⁡x→0+1cos⁡(9x)=lim⁡x→0+(18xsin⁡(9x))\lim_{x \to 0^+}\left(\frac{1}{8 x} \sin{\left(9 x \right)}\right) \lim_{x \to 0^+} \frac{1}{\cos{\left(9 x \right)}} = \lim_{x \to 0^+}\left(\frac{1}{8 x} \sin{\left(9 x \right)}\right)
Do replacement
u=9xu = 9 x
then
lim⁡x→0+(sin⁡(9x)8x)=lim⁡u→0+(9sin⁡(u)8u)\lim_{x \to 0^+}\left(\frac{\sin{\left(9 x \right)}}{8 x}\right) = \lim_{u \to 0^+}\left(\frac{9 \sin{\left(u \right)}}{8 u}\right)
=
9lim⁡u→0+(sin⁡(u)u)8\frac{9 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{8}
The limit
lim⁡u→0+(sin⁡(u)u)\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)
is first remarkable limit, is equal to 1.

The final answer:
lim⁡x→0+(tan⁡(9x)8x)=98\lim_{x \to 0^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) = \frac{9}{8}
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
lim⁡x→0+tan⁡(9x)=0\lim_{x \to 0^+} \tan{\left(9 x \right)} = 0
and limit for the denominator is
lim⁡x→0+(8x)=0\lim_{x \to 0^+}\left(8 x\right) = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(tan⁡(9x)8x)\lim_{x \to 0^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right)
=
Let's transform the function under the limit a few
lim⁡x→0+(tan⁡(9x)8x)\lim_{x \to 0^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right)
=
lim⁡x→0+(ddxtan⁡(9x)ddx8x)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan{\left(9 x \right)}}{\frac{d}{d x} 8 x}\right)
=
lim⁡x→0+(9tan⁡2(9x)8+98)\lim_{x \to 0^+}\left(\frac{9 \tan^{2}{\left(9 x \right)}}{8} + \frac{9}{8}\right)
=
lim⁡x→0+(9tan⁡2(9x)8+98)\lim_{x \to 0^+}\left(\frac{9 \tan^{2}{\left(9 x \right)}}{8} + \frac{9}{8}\right)
=
98\frac{9}{8}
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
02468-8-6-4-2-101020-10
Plot the graph
Rapid solution [src] 
9/8
98\frac{9}{8}
Expand and simplify
One‐sided limits [src] 
     /tan(9*x)\
 lim |--------|
x->0+\  8*x   /
lim⁡x→0+(tan⁡(9x)8x)\lim_{x \to 0^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) 
9/8
98\frac{9}{8} 
= 1.125
     /tan(9*x)\
 lim |--------|
x->0-\  8*x   /
lim⁡x→0−(tan⁡(9x)8x)\lim_{x \to 0^-}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) 
9/8
98\frac{9}{8} 
= 1.125
= 1.125
Other limits x→0, -oo, +oo, 1
lim⁡x→0−(tan⁡(9x)8x)=98\lim_{x \to 0^-}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) = \frac{9}{8}
More at x→0 from the left
lim⁡x→0+(tan⁡(9x)8x)=98\lim_{x \to 0^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) = \frac{9}{8}
lim⁡x→∞(tan⁡(9x)8x)\lim_{x \to \infty}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right)
More at x→oo
lim⁡x→1−(tan⁡(9x)8x)=tan⁡(9)8\lim_{x \to 1^-}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) = \frac{\tan{\left(9 \right)}}{8}
More at x→1 from the left
lim⁡x→1+(tan⁡(9x)8x)=tan⁡(9)8\lim_{x \to 1^+}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right) = \frac{\tan{\left(9 \right)}}{8}
More at x→1 from the right
lim⁡x→−∞(tan⁡(9x)8x)\lim_{x \to -\infty}\left(\frac{\tan{\left(9 x \right)}}{8 x}\right)
More at x→-oo
Numerical answer [src] 
1.125
1.125
The graph
Limit of the function tan(9*x)/(8*x)
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