Find the limit limx→π2(cos⁡(5x)cos⁡(3x)) without using L'Hospital's rule

cos⁡(5x)=sin⁡(52π−5x)=sin⁡5(π2−x)Andcos⁡(3x)=−sin⁡(32π−3x)=−sin⁡3(π2−x)So we set π2−x=was x→π2 we have x→0The given limit can be written aslimw→0sin⁡5w−sin⁡3w=−53limw→03wsin⁡5w5wsin⁡3w=−53limw→0(sin⁡5w5w⋅3wsin⁡3w)=−53Hope this can be useful

Write&nbsp;t=x−π2, then&nbsp;limx→π2cos⁡(5x)cos⁡(3x)=limt→0sin⁡(−5t−2π)sin⁡(−3t−π)=limt→0−sin⁡(5t)sin⁡(3t)=⋅limt→0−sin⁡(5t)5t⋅3tsin⁡(3t)⋅53=−53⋅limt→0sin⁡(5t)5t⋅limt→03tsin⁡(3t)=-53·limt→0sin(5t)5t⏟=1·(limt→0sin(3t)3t⏟=1)-1=−53

cos⁡(5x)=sin⁡(52π−5x)=sin⁡5(π2−x)Andcos⁡(3x)=−sin⁡(32π−3x)=−sin⁡3(π2−x)So we set π2−x=was x→π2 we have x→0The given limit can be written aslimw→0sin⁡5w−sin⁡3w=−53limw→03wsin⁡5w5wsin⁡3w=−53limw→0(sin⁡5w5w⋅3wsin⁡3w)=−53Hope this can be useful

Write&nbsp;t=x−π2, then&nbsp;limx→π2cos⁡(5x)cos⁡(3x)=limt→0sin⁡(−5t−2π)sin⁡(−3t−π)=limt→0−sin⁡(5t)sin⁡(3t)=⋅limt→0−sin⁡(5t)5t⋅3tsin⁡(3t)⋅53=−53⋅limt→0sin⁡(5t)5t⋅limt→03tsin⁡(3t)=-53·limt→0sin(5t)5t⏟=1·(limt→0sin(3t)3t⏟=1)-1=−53

Key to "Find the limit limx→π2(cos⁡(5x)cos⁡(3x)) without using L'Hospital's rule"

Cash Duncan

Answered question

2022-04-14

Find the limit

lim_{x \to \frac{π}{2}} (\frac{\cos (5 x)}{\cos (3 x)})

without using L'Hospital's rule

Answer & Explanation

Wernbergbo9d

Beginner2022-04-15Added 12 answers

\cos (5 x) = \sin (\frac{5}{2} π - 5 x) = \sin 5 (\frac{π}{2} - x)

And

\cos (3 x) = - \sin (\frac{3}{2} π - 3 x) = - \sin 3 (\frac{π}{2} - x)

So we set

\frac{π}{2} - x = w

x \to \frac{π}{2}

we have

x \to 0

The given limit can be written as

lim_{w \to 0} \frac{\sin 5 w}{- \sin 3 w} = - \frac{5}{3} lim_{w \to 0} \frac{3 w \sin 5 w}{5 w \sin 3 w} = - \frac{5}{3} lim_{w \to 0} (\frac{\sin 5 w}{5 w} \cdot \frac{3 w}{\sin 3 w}) = - \frac{5}{3}

Hope this can be useful

resacarno4u

Beginner2022-04-16Added 12 answers

Write $t = x - \frac{π}{2}$ , then
$lim_{x \to \frac{π}{2}} \frac{\cos (5 x)}{\cos (3 x)} = lim_{t \to 0} \frac{\sin (- 5 t - 2 π)}{\sin (- 3 t - π)}$
$= lim_{t \to 0} \frac{- \sin (5 t)}{\sin (3 t)}$
$= \cdot lim_{t \to 0} \frac{- \sin (5 t)}{5 t} \cdot \frac{3 t}{\sin (3 t)} \cdot \frac{5}{3}$
$= - \frac{5}{3} \cdot lim_{t \to 0} \frac{\sin (5 t)}{5 t} \cdot lim_{t \to 0} \frac{3 t}{\sin (3 t)}$
$= - \frac{5}{3} \cdot \underset{= 1}{\underset{⏟}{\lim_{t \to 0} \frac{\sin (5 t)}{5 t}}} \cdot (\underset{= 1}{\underset{⏟}{\lim_{t \to 0} \frac{\sin (3 t)}{3 t}}})^{- 1}$
$= - \frac{5}{3}$

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