# Math Help - Trig limit

1. ## Trig limit

Evaluate the following limit without using l'Hopitals rule.
$\lim_{x\to\frac{\pi}{4}}\frac{6-6tanx}{sinx-cosx}$

No matter how I attempt to manipulate it I always end up with a 0 in the denominator, if anyone could help?

2. Originally Posted by xxlvh
Evaluate the following limit without using l'Hopitals rule.
$\lim_{x\to\frac{\pi}{4}}\frac{6-6tanx}{sinx-cosx}$

No matter how I attempt to manipulate it I always end up with a 0 in the denominator, if anyone could help?
$\frac{6-6\tan{x}}{\sin{x}-\cos{x}} =$

$\frac{6(1-\tan{x})}{\sin{x}-\cos{x}} =$

$\frac{6\left(\frac{\cos{x}}{\cos{x}} -\frac{\sin{x}}{\cos{x}}\right)}{\sin{x}-\cos{x}} =$

$\frac{6\left(\frac{\cos{x}-\sin{x}}{\cos{x}}\right)}{\sin{x}-\cos{x}} =$

$\frac{6(\cos{x}-\sin{x})}{\cos{x}(\sin{x}-\cos{x})} =$

$-\frac{6}{\cos{x}}$

now find the limit