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

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

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     /     /1\  \
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 lim |----------|
x->oo|   /  1  \|
     |tan|-----||
     \   \1 + x//
lim⁡x→∞(tan⁡(1x)tan⁡(1x+1))\lim_{x \to \infty}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right) 
Limit(tan(1/x)/tan(1/(1 + x)), x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
lim⁡x→∞1tan⁡(1x+1)=∞\lim_{x \to \infty} \frac{1}{\tan{\left(\frac{1}{x + 1} \right)}} = \infty
and limit for the denominator is
lim⁡x→∞1tan⁡(1x)=∞\lim_{x \to \infty} \frac{1}{\tan{\left(\frac{1}{x} \right)}} = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→∞(tan⁡(1x)tan⁡(1x+1))\lim_{x \to \infty}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right)
=
lim⁡x→∞(ddx1tan⁡(1x+1)ddx1tan⁡(1x))\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{\tan{\left(\frac{1}{x + 1} \right)}}}{\frac{d}{d x} \frac{1}{\tan{\left(\frac{1}{x} \right)}}}\right)
=
lim⁡x→∞(x2(tan⁡2(1x+1)+1)tan⁡2(1x)(x+1)2(tan⁡2(1x)+1)tan⁡2(1x+1))\lim_{x \to \infty}\left(\frac{x^{2} \left(\tan^{2}{\left(\frac{1}{x + 1} \right)} + 1\right) \tan^{2}{\left(\frac{1}{x} \right)}}{\left(x + 1\right)^{2} \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right) \tan^{2}{\left(\frac{1}{x + 1} \right)}}\right)
=
lim⁡x→∞(1(1tan⁡2(1x)+2xtan⁡2(1x)+1x2tan⁡2(1x))tan⁡2(1x+1))\lim_{x \to \infty}\left(\frac{1}{\left(\frac{1}{\tan^{2}{\left(\frac{1}{x} \right)}} + \frac{2}{x \tan^{2}{\left(\frac{1}{x} \right)}} + \frac{1}{x^{2} \tan^{2}{\left(\frac{1}{x} \right)}}\right) \tan^{2}{\left(\frac{1}{x + 1} \right)}}\right)
=
lim⁡x→∞(ddx1tan⁡2(1x+1)ddx(1tan⁡2(1x)+2xtan⁡2(1x)+1x2tan⁡2(1x)))\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{\tan^{2}{\left(\frac{1}{x + 1} \right)}}}{\frac{d}{d x} \left(\frac{1}{\tan^{2}{\left(\frac{1}{x} \right)}} + \frac{2}{x \tan^{2}{\left(\frac{1}{x} \right)}} + \frac{1}{x^{2} \tan^{2}{\left(\frac{1}{x} \right)}}\right)}\right)
=
lim⁡x→∞(2(tan⁡2(1x+1)+1)(x+1)2(2(tan⁡2(1x)+1)x2tan⁡3(1x)−2x2tan⁡2(1x)+4(tan⁡2(1x)+1)x3tan⁡3(1x)−2x3tan⁡2(1x)+2(tan⁡2(1x)+1)x4tan⁡3(1x))tan⁡3(1x+1))\lim_{x \to \infty}\left(\frac{2 \left(\tan^{2}{\left(\frac{1}{x + 1} \right)} + 1\right)}{\left(x + 1\right)^{2} \left(\frac{2 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{2} \tan^{3}{\left(\frac{1}{x} \right)}} - \frac{2}{x^{2} \tan^{2}{\left(\frac{1}{x} \right)}} + \frac{4 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{3} \tan^{3}{\left(\frac{1}{x} \right)}} - \frac{2}{x^{3} \tan^{2}{\left(\frac{1}{x} \right)}} + \frac{2 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{4} \tan^{3}{\left(\frac{1}{x} \right)}}\right) \tan^{3}{\left(\frac{1}{x + 1} \right)}}\right)
=
lim⁡x→∞(2(x+1)2(2(tan⁡2(1x)+1)x2tan⁡3(1x)−2x2tan⁡2(1x)+4(tan⁡2(1x)+1)x3tan⁡3(1x)−2x3tan⁡2(1x)+2(tan⁡2(1x)+1)x4tan⁡3(1x))tan⁡3(1x+1))\lim_{x \to \infty}\left(\frac{2}{\left(x + 1\right)^{2} \left(\frac{2 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{2} \tan^{3}{\left(\frac{1}{x} \right)}} - \frac{2}{x^{2} \tan^{2}{\left(\frac{1}{x} \right)}} + \frac{4 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{3} \tan^{3}{\left(\frac{1}{x} \right)}} - \frac{2}{x^{3} \tan^{2}{\left(\frac{1}{x} \right)}} + \frac{2 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{4} \tan^{3}{\left(\frac{1}{x} \right)}}\right) \tan^{3}{\left(\frac{1}{x + 1} \right)}}\right)
=
lim⁡x→∞(2(x+1)2(2(tan⁡2(1x)+1)x2tan⁡3(1x)−2x2tan⁡2(1x)+4(tan⁡2(1x)+1)x3tan⁡3(1x)−2x3tan⁡2(1x)+2(tan⁡2(1x)+1)x4tan⁡3(1x))tan⁡3(1x+1))\lim_{x \to \infty}\left(\frac{2}{\left(x + 1\right)^{2} \left(\frac{2 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{2} \tan^{3}{\left(\frac{1}{x} \right)}} - \frac{2}{x^{2} \tan^{2}{\left(\frac{1}{x} \right)}} + \frac{4 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{3} \tan^{3}{\left(\frac{1}{x} \right)}} - \frac{2}{x^{3} \tan^{2}{\left(\frac{1}{x} \right)}} + \frac{2 \left(\tan^{2}{\left(\frac{1}{x} \right)} + 1\right)}{x^{4} \tan^{3}{\left(\frac{1}{x} \right)}}\right) \tan^{3}{\left(\frac{1}{x + 1} \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) 2 time(s)
The graph
02468-8-6-4-2-1010-5050
Plot the graph
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(tan⁡(1x)tan⁡(1x+1))=1\lim_{x \to \infty}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right) = 1
lim⁡x→0−(tan⁡(1x)tan⁡(1x+1))\lim_{x \to 0^-}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right)
More at x→0 from the left
lim⁡x→0+(tan⁡(1x)tan⁡(1x+1))\lim_{x \to 0^+}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right)
More at x→0 from the right
lim⁡x→1−(tan⁡(1x)tan⁡(1x+1))=tan⁡(1)tan⁡(12)\lim_{x \to 1^-}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right) = \frac{\tan{\left(1 \right)}}{\tan{\left(\frac{1}{2} \right)}}
More at x→1 from the left
lim⁡x→1+(tan⁡(1x)tan⁡(1x+1))=tan⁡(1)tan⁡(12)\lim_{x \to 1^+}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right) = \frac{\tan{\left(1 \right)}}{\tan{\left(\frac{1}{2} \right)}}
More at x→1 from the right
lim⁡x→−∞(tan⁡(1x)tan⁡(1x+1))=1\lim_{x \to -\infty}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right) = 1
More at x→-oo
Rapid solution [src] 
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11
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Limit of the function tan(1/x)/tan(1/(1+x))
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