Cos Sine to the tan limit

Calculus Level 2

$\large \lim_{ x \to 0} \left(\frac{ \cos ^4 (3x) + \cos x}{2\cos (2x)} \right) ^{\frac{1}{\tan (x)}} = \ ?$

The answer is 1.

2 solutions

Rohit Ner
Jun 20, 2015

$\large \begin{aligned}\lim_{ x \to 0} \left(\frac{ \cos ^4 3x + \cos x}{2\cos 2x} \right) ^{\frac{1}{\tan x}} & \large={ e }^{ \displaystyle\lim _{ x\to 0 } \left( \frac { \cos ^{ 4 } 3x+\cos x }{ 2\cos 2x } -1 \right) \left( \frac { 1 }{ \tan { x } } \right) } \\ & \large={ e }^{ \displaystyle\lim _{ x\to 0 } \left( \frac { 1+1 }{ 2 } -1 \right) \left( \frac { 1 }{ \tan { x } } \right) } \\ & \large={ e }^{ \displaystyle\lim _{ x\to 0 } \left( \frac { 0 }{ \tan { x } } \right) } \\ & \large={e}^{0} \\ & \large \Huge\color{#3D99F6}{=\boxed 1} \end{aligned}$

${ e }^{ \displaystyle\lim _{ x\to 0 } \left( \frac { 0 }{ \tan { x } } \right) }$ The term stated above is not a $\frac{0}{0}$ form because the value of $x$ tends to zero and doesnt exactly correspond to zero.

Moderator note:

Your very first line already don't make sense.

${ e }^{ \lim _{ x\to 0 } \left( \frac { \cos ^{ 4 } 3x+\cos x }{ 2\cos 2x } -1 \right) \left( \frac { 1 }{ \tan { x } } \right) }$ $\neq { e }^{ \lim _{ x\to 0 } \left( \frac { 1+1 }{ 2 } -1 \right) \left( \frac { 1 }{ \tan { x } } \right) }$ $={ e }^{ \lim _{ x\to 0 } \left( \frac { \rightarrow 1+\rightarrow 1 }{ \rightarrow 2 } -1 \right) \left( \frac { 1 }{ \tan { x } } \right) }$

You did it wrong.

Abhishek Sharma - 5 years, 11 months ago

Yes you are right. Further simplification of the first bracket in you comment tends to zero. So what's wrong with my solution?

Rohit Ner - 5 years, 11 months ago

Correct and complete your solution.

Abhishek Sharma - 5 years, 11 months ago

@Abhishek Sharma – Actually the steps you mentioned are too obvious and don't need to be necessarily mentioned. :)

Rohit Ner - 5 years, 11 months ago

@Rohit Ner – So what is the point of writing the solution?

Abhishek Sharma - 5 years, 11 months ago

@Abhishek Sharma – My solution just intends to give a idea of the approach. You can post yours if you find mine not up to the mark.

Rohit Ner - 5 years, 11 months ago

Nice solution rohit.Absolutely perfect.

shivansh mittal - 5 years, 11 months ago

pls explain the 1st line . how that -1 came ? i want to knw more about limits . plzz help me

Anand O R - 5 years, 11 months ago

Shivansh Mittal
Jun 20, 2015

Put x=0 →cos0=1 , (cos3×0)^4=1 ,cos2×0=1 and in 1/tanx would be tending to infinity. We know that (1)^→infinity =1. Hence Ans =1.

Moderator note:

Your answer is wrong. $1^\infty$ is an indeterminate form.

You got that answer by luck. $1^\infty$ is indeterminate form.

Kushal Patankar - 5 years, 11 months ago

Absolute 1^tending toinfinity is not an indeterminate form.Infact tending to 1^ tending to infinity is an indeterminate form

shivansh mittal - 5 years, 11 months ago

Kushal kindly revisit the indeterminate forms before replying.The following are indeterminate forms:- 1)→0/→0 2)→infinity/→infinity 3)→infinity-→infinity 4)→infinity ×→0 5)(→1)^→infinity 6)(→0)^→0 7)(→infinity)^→0 And except them no other indeterminate form exists.

shivansh mittal - 5 years, 11 months ago

Cos Sine to the tan limit

The answer is 1.

2 solutions

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