Limit of $lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$ (No L'Hôpital)
$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
begin{align*}
& lim_{x to 0} frac{cos 2x}{sin 2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
lim_{x to 0}frac{frac{cos2x }{2x}}{frac{sin 2x}{2x}}cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} frac{cos 2x}{2x} cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} frac{{cos^2 (x)}-{sin^2 (x)}}{2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{cos^2(x)}{2x}-frac{sin^2 x}{2x}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1-sin^2 x}{2x}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-frac{sin^2 x}{2x}-frac{sin x}{2}right) cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-frac{sin x}{2}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-2frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-sin xright)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
calculus limits trigonometry limits-without-lhopital
edited Nov 24 at 18:57
quid♦
36.9k95093
asked Nov 24 at 13:14
Jakcjones
318
add a comment |
$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
begin{align*}
& lim_{x to 0} frac{cos 2x}{sin 2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
lim_{x to 0}frac{frac{cos2x }{2x}}{frac{sin 2x}{2x}}cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} frac{cos 2x}{2x} cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} frac{{cos^2 (x)}-{sin^2 (x)}}{2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{cos^2(x)}{2x}-frac{sin^2 x}{2x}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1-sin^2 x}{2x}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-frac{sin^2 x}{2x}-frac{sin x}{2}right) cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-frac{sin x}{2}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-2frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-sin xright)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
calculus limits trigonometry limits-without-lhopital
edited Nov 24 at 18:57
quid♦
36.9k95093
asked Nov 24 at 13:14
Jakcjones
318
add a comment |
$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
begin{align*}
& lim_{x to 0} frac{cos 2x}{sin 2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
lim_{x to 0}frac{frac{cos2x }{2x}}{frac{sin 2x}{2x}}cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} frac{cos 2x}{2x} cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} frac{{cos^2 (x)}-{sin^2 (x)}}{2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{cos^2(x)}{2x}-frac{sin^2 x}{2x}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1-sin^2 x}{2x}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-frac{sin^2 x}{2x}-frac{sin x}{2}right) cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-frac{sin x}{2}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-2frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-sin xright)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
calculus limits trigonometry limits-without-lhopital
edited Nov 24 at 18:57
quid♦
36.9k95093
asked Nov 24 at 13:14
Jakcjones
318
$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
begin{align*}
& lim_{x to 0} frac{cos 2x}{sin 2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
lim_{x to 0}frac{frac{cos2x }{2x}}{frac{sin 2x}{2x}}cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} frac{cos 2x}{2x} cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} frac{{cos^2 (x)}-{sin^2 (x)}}{2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{cos^2(x)}{2x}-frac{sin^2 x}{2x}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1-sin^2 x}{2x}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-frac{sin^2 x}{2x}-frac{sin x}{2}right) cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-frac{sin x}{2}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-2frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-sin xright)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
calculus limits trigonometry limits-without-lhopital
calculus limits trigonometry limits-without-lhopital
edited Nov 24 at 18:57
quid♦
36.9k95093
asked Nov 24 at 13:14
Jakcjones
318
edited Nov 24 at 18:57
quid♦
36.9k95093
asked Nov 24 at 13:14
Jakcjones
318
edited Nov 24 at 18:57
quid♦
36.9k95093
edited Nov 24 at 18:57
quid♦
36.9k95093
edited Nov 24 at 18:57
quid♦
36.9k95093
36.9k95093
asked Nov 24 at 13:14
Jakcjones
318
asked Nov 24 at 13:14
Jakcjones
318
asked Nov 24 at 13:14
Jakcjones
318
318
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
HINT
We have that
$$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$
answered Nov 24 at 13:19
gimusi
1
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
– gimusi
Nov 24 at 13:42
add a comment |
Hint:
$$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$
answered Nov 24 at 13:18
Nosrati
26.4k62353
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:28
add a comment |
$displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$
$displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $
answered Nov 24 at 13:20
Yadati Kiran
1,702519
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
add a comment |
Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
$$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$
answered Nov 24 at 13:17
greedoid
37.5k114794
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:26
add a comment |
$$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$
For $tan xne0,F(x)=dfrac{1-tan^2x}2$
As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$
answered Nov 26 at 6:49
lab bhattacharjee
222k15156274
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
We have that
$$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$
answered Nov 24 at 13:19
gimusi
1
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
– gimusi
Nov 24 at 13:42
add a comment |
HINT
We have that
$$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$
answered Nov 24 at 13:19
gimusi
1
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
– gimusi
Nov 24 at 13:42
add a comment |
HINT
We have that
$$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$
answered Nov 24 at 13:19
gimusi
1
HINT
We have that
$$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$
answered Nov 24 at 13:19
gimusi
1
edited Nov 24 at 13:46
answered Nov 24 at 13:19
gimusi
1
answered Nov 24 at 13:19
gimusi
1
answered Nov 24 at 13:19
gimusi
1
1
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
– gimusi
Nov 24 at 13:42
add a comment |
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
– gimusi
Nov 24 at 13:42
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
– gimusi
Nov 24 at 13:42
Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
– gimusi
Nov 24 at 13:42
add a comment |
Hint:
$$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$
answered Nov 24 at 13:18
Nosrati
26.4k62353
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:28
add a comment |
Hint:
$$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$
answered Nov 24 at 13:18
Nosrati
26.4k62353
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:28
add a comment |
Hint:
$$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$
answered Nov 24 at 13:18
Nosrati
26.4k62353
Hint:
$$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$
answered Nov 24 at 13:18
Nosrati
26.4k62353
answered Nov 24 at 13:18
Nosrati
26.4k62353
answered Nov 24 at 13:18
Nosrati
26.4k62353
answered Nov 24 at 13:18
Nosrati
26.4k62353
26.4k62353
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:28
add a comment |
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:28
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:28
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:28
add a comment |
$displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$
$displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $
answered Nov 24 at 13:20
Yadati Kiran
1,702519
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
add a comment |
$displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$
$displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $
answered Nov 24 at 13:20
Yadati Kiran
1,702519
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
add a comment |
$displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$
$displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $
answered Nov 24 at 13:20
Yadati Kiran
1,702519
$displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$
$displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $
answered Nov 24 at 13:20
Yadati Kiran
1,702519
answered Nov 24 at 13:20
Yadati Kiran
1,702519
answered Nov 24 at 13:20
Yadati Kiran
1,702519
answered Nov 24 at 13:20
Yadati Kiran
1,702519
1,702519
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
add a comment |
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:27
add a comment |
Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
$$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$
answered Nov 24 at 13:17
greedoid
37.5k114794
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:26
add a comment |
Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
$$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$
answered Nov 24 at 13:17
greedoid
37.5k114794
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:26
add a comment |
Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
$$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$
answered Nov 24 at 13:17
greedoid
37.5k114794
Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
$$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$
answered Nov 24 at 13:17
greedoid
37.5k114794
answered Nov 24 at 13:17
greedoid
37.5k114794
answered Nov 24 at 13:17
greedoid
37.5k114794
answered Nov 24 at 13:17
greedoid
37.5k114794
37.5k114794
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:26
add a comment |
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:26
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:26
Thanks, that tan(x) identity was what saved me :D
– Jakcjones
Nov 24 at 13:26
add a comment |
$$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$
For $tan xne0,F(x)=dfrac{1-tan^2x}2$
As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$
answered Nov 26 at 6:49
lab bhattacharjee
222k15156274
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$$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$
For $tan xne0,F(x)=dfrac{1-tan^2x}2$
As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$
answered Nov 26 at 6:49
lab bhattacharjee
222k15156274
add a comment |
$$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$
For $tan xne0,F(x)=dfrac{1-tan^2x}2$
As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$
answered Nov 26 at 6:49
lab bhattacharjee
222k15156274
$$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$
For $tan xne0,F(x)=dfrac{1-tan^2x}2$
As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$
answered Nov 26 at 6:49
lab bhattacharjee
222k15156274
answered Nov 26 at 6:49
lab bhattacharjee
222k15156274
answered Nov 26 at 6:49
lab bhattacharjee
222k15156274
answered Nov 26 at 6:49
lab bhattacharjee
222k15156274
222k15156274
add a comment |
add a comment |
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