Can someone explain this trigonometric limit without L'Hopital? I can not solve this limit:
$$\lim \limits_{x\to 0}\frac{x^2}{1-\sec(x)}$$
$$\lim \limits_{x\to 0}  \frac{x^2}{1-\sec(x)}=\lim \limits_{x\to 0}\frac {x^2}{1-\sec(x)}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{1-\sec^2(x)}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{-\tan^2(x)}$$
 A: $\lim \limits_{x\to 0}  \frac{x^2}{1-\sec x}= \lim \limits_{x\to 0} \frac {x^2}{1-\sec x}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{\cos^2 x-1}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{-\sin^2(x)}$
Now if you know the limit of $\frac {\sin x}x$ the other terms are well behaved.
A: Note:
I fixed an error noted by triple_sec .
$\dfrac{x^2}{1-\sec x}
=\dfrac{x^2}{1-1/\cos x}
=\dfrac{ x^2 \cos x}{\cos x-1}
$.
Using
$\cos(2x)
=\cos^2(x)-\sin^2(x)
=1-2\sin^2(x)
$,
$\cos(x)-1
=-2\sin^2(x/2)
$
(I originally had $+$ here instead of $-$)
so
$\dfrac{x^2}{1-\sec x}
=\dfrac{ x^2 \cos x}{\cos x-1}
=\dfrac{ x^2 \cos x}{-2\sin^2(x/2)}
=- \cos x\dfrac{ x^2 }{-2\sin^2(x/2)}
$.
Since,
as $x \to 0$,
$\cos x \to 1$
and
$\dfrac{\sin x}{x}
\to 1$,
$-\cos x\dfrac{ x^2}{2\sin^2(x/2)}
=-2\cos x\left(\dfrac{ x/2 }{\sin(x/2)}\right)^2
\to -2
$.
A: Hint: Multiply by $\cos x$ on the top and the bottom, and then multiply the top and the bottom by $\cos x +1$ and use the identity $1-\cos^{2}x=\sin^{2}x$.
A: What are you allowed to use? If Maclaurin series are OK, use $\cos x = 1 -\frac{x^2}{2} + O(x^4)$, hence your expression after a bit of algebra becomes
$$
\lim_{x \to 0}(1-\frac{x^2}{2} +O(x^4))\lim_{x \to 0}\frac{x^2}{-\frac{x^2}{2}+O(x^4)}=1 \cdot (-2)=-2
$$
