Question

8. Find the value of lim = o(Sin(2x))^Tan_ (2x)?​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{lim_{x\to0}\{sin(2x)\}^{tan(2x)}}$

$\sf{\purple{Let\,\,\,2x=t}}\\\sf{\purple{so, \,\,\,as\,\,\,x\,\to\,0\,\,\implies\,t\,\to\,0}}$

So, the limit becomes,

$\sf{lim_{t\to0}\{sin(t)\}^{tan(t)}}$

$\sf{=lim_{t\to0}(e)^{tan(t)\,ln(sin(t))}}$

$\sf{=lim_{t\to0}(e)^{\dfrac{ln(sin(t))}{cot(t)}}}$

$\sf{=(e)^{lim_{t\to0}\dfrac{ln(sin(t))}{cot(t)}}}$

This is 0/0 form, so, using l'hospital's rule,

$\sf{=(e)^{lim_{t\to0}\dfrac{cot(t)}{-cosec^2(t)}}}$

$\sf{=(e)^{-\,lim_{t\to0}\dfrac{cos(t)sin^2(t)}{sin(t)}}}$

$\sf{=(e)^{-\,lim_{t\to0}\,cos(t)sin(t)}}$

$\sf{=(e)^{-\,0}}$

$\sf{=e^{0}}$

$\sf{=1}$