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What is the limit as x=>0 of sin(x)/x+tan(x) ?

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hinkberries | Student, Undergraduate | (Level 1) Honors

Posted September 30, 2010 at 6:23 AM via web

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What is the limit as x=>0 of sin(x)/x+tan(x) ?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted September 30, 2010 at 3:05 PM (Answer #1)

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It is not clear if the denominator of the ratio is just x or is (x+tan x).

This thing must be specified with the help of the brackets.

If the denominator of the ratio is x:

lim [sin(x)/x+tan(x)] = lim [sin(x)/x] + lim tan(x)

lim [sin(x)/x] is an elementary limit and the result is1.

We'll calculate lim tan x.

We'll substitute x by the value 0.

lim tan x = tan 0 = 0

So, the limit is:

lim [sin(x)/x+tan(x)] = 1 + 0 = 1

 

Now, if the denominator is (x+tan x), we'll calculate the limit:

lim [sin(x)/(x+tanx)]  = sin 0/(x + tan 0)

lim [sin(x)/(x+tanx)]  = 0/0

"0/0" is an indetermination

We'll use l'hospital rule. We'll differentiate separately numerator and denominator.

(sinx)' = cos x

(x+tanx)' = 1 + 1/(cos x)^2

lim [sin(x)/(x+tanx)]  = lim (sinx)'/(x+tanx)'

lim (sinx)'/(x+tanx)' = lim cos x/[1 + 1/(cos x)^2]

We'll substitute x by 0:

lim cos x/[1 + 1/(cos x)^2] =cos 0/[1 + 1/(cos 0)^2]

cos 0/[1 + 1/(cos 0)^2] = 1/(1+1) = 1/2

lim [sin(x)/(x+tanx)]  = 1/2

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william1941 | College Teacher | (Level 3) Valedictorian

Posted September 30, 2010 at 10:31 AM (Answer #2)

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We have to find the value of  sin (x) / x + tan (x) for lim x-->0.

We can use the Taylor series for sin x

So sin x = x - x^3/3! + x^5/5! ...

=> sin x / x = 1- x^2/ 3! + x^4/ 5! ...

tan x =  x + x^3/3 + 2x^5/15 +...

So for lim x--->0

sin x / x = 1- x^2/ 3! + x^4/ 5! ... = 1

and tan x = 0

The expression sin (x) / x + tan (x) for lim x-->0 = 1.

Therefore the result for sin (x) / x + tan (x) for lim x-->0 = 1.

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neela | High School Teacher | (Level 3) Valedictorian

Posted September 30, 2010 at 12:12 PM (Answer #3)

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The limit of sinx/x+tanx, as x-->0.

Solution:

The most difficult symbol of  slash "/"

Do you mean  (i) sinx/x+tanx = [(sinx)/x ]+tanx  or  (ii) (sinx)/(x+tanx)?

We answer both cases.

(i)

The limit of sinx/x + tanx = lt sinx/x +lt tanx = 1+0 =1.

The second term lttanx is got by just by substitution  of  x=0 in tanx which gives tan0 = 0.

The lt of sinx/x is of special interest , as  putting 0 in (sinx)/x  becomes sin0/0  which is a 0/0 form of indetermination.

So we can use L'Hospital's rule of diffrentiating the numerator and denominator and then take the limit. Or we can go for the geometrical proof where (sinx/x) remains between cosx and 1 as x  approaches zero. But cosx approaches 1. So Limit (sinx)/x should also approach 1.

Lt of (sinx)/x by L'Hospital's rule :

Lt x--> 0  (sinx)/x = Lt x-->0  (sinx)'/ (x)' = Lt x-->0 (cosx)/1 = cos0 = 1.

So lt (sinx)/x = 1.

Therefore Lt {(sinx)/x +tanx} =  1+0 = 1. 

(ii) To find the lt x--> 0 sinx /(x+tanx);

We know already (as in (i) ) that Lt  x--> 0 (sinx)/x = 1

Simlarly lt x-->0 (tanx)/x =  Lt (sinx)/(xcosx) = 1/cos0 = 1.

Therefore lt x--> 0 (sinx)/(x+tanx) = lt {[(sinx)/x]/[(x+tanx)/x]}. Here we devided both numerator and denominator by x.

Lt x--> 0 (sinx)/(x+tanx) =  {lt x-->0 (Sinx)/x}/ {lt x-->0 [(x/x)+(tanx)/x]}

Lt x-->0 (sinx)/(x+tanx) = 1/(1+1) = 1/2

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