1. ## trig limit

Find the value of this limit using the fastest non-l'Hopital method...

$\displaystyle \displaystyle\lim_{x\to{0}}\frac{x^2sec\;x}{sec\;x-1}$

Thanks!

2. Originally Posted by polymerase
$\displaystyle \displaystyle\lim_{x\to{0}}\frac{x^2sec\;x}{sec\;x-1}$

Thanks!
You may assume that $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12.$

Now, multiply top & bottom by $\displaystyle \cos x.$

3. $\displaystyle \lim_{x\rightarrow{0}}{\frac{x^2\cdot{\sec(x)}}{\s ec(x)-1}}=\lim_{x\rightarrow{0}}{\frac{x^2\cdot{\frac{1} {\cos(x)}}}{\frac{1}{\cos(x)}-1}}$

THen: $\displaystyle \lim_{x\rightarrow{0}}{\frac{x^2\cdot{\sec(x)}}{\s ec(x)-1}}=\lim_{x\rightarrow{0}}{\frac{x^2}{1-\cos(x)}}$

Now we multiply numerator and denominator by $\displaystyle 1+\cos(x)$

We get: $\displaystyle \lim_{x\rightarrow{0}}{\frac{x^2\cdot{\sec(x)}}{\s ec(x)-1}}=\lim_{x\rightarrow{0}}{\frac{x^2(1+\cos(x))}{1-\cos^2(x)}}$

But $\displaystyle \sin^2(x)=1-\cos^2(x)$ and $\displaystyle (1+\cos(x))\rightarrow 2$ so: $\displaystyle \lim_{x\rightarrow{0}}{\frac{x^2\cdot{\sec(x)}}{\s ec(x)-1}}=2\lim_{x\rightarrow{0}}{\frac{x^2}{\sin^2(x)}} =2$

PS- Krizalid is faster than me

4. Just note that:$\displaystyle \frac{{x^2 \sec (x)}}{{\sec (x) - 1}} = \frac{{x^2 }}{{1 - \cos (x)}} = \frac{{x^2 \left( {1 + \cos (x)} \right)}}{{\sin ^2 (x)}}$.