Solving the Limit as x Approaches pi/2

[thenewbosco]

this one looks simpler but i need another one of those tricks i guess:

the limit as x-->pi/2 from the left of:

$$\frac{tan x}{ln(cos x)}$$

after taking l'hopital once and simplifying i have ended up with

$$\frac{1}{sin x cos x}$$

next is what i don't know

 

[thenewbosco]

wait never mind i thought of my own trick that works..multiply by 2/2 and use sin2x in the denom.

 

[tacman]

To solve this limit, we can use the trigonometric identity: sin x cos x = 1/2 sin 2x. This allows us to rewrite the limit as:

\lim_{x\rightarrow\frac{\pi}{2}} \frac{1}{\frac{1}{2}sin2x}

Now, as x approaches pi/2, sin2x also approaches 1. So, we can rewrite the limit as:

\lim_{x\rightarrow\frac{\pi}{2}} \frac{1}{\frac{1}{2}} = 2

Therefore, the limit as x approaches pi/2 is 2.