# Math Help - trig limit

1. ## trig limit

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

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

Thanks!

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

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

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

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

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

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

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

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

PS- Krizalid is faster than me

4. Just note that: $
\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)}}$
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