Can someone explain this trigonometric limit without L'Hopital?

Question

I can not solve this limit: $$\lim \limits_{x\to 0}\frac{x^2}{1-\sec(x)}$$

$$\lim \limits_{x\to 0} \frac{x^2}{1-\sec(x)}=\lim \limits_{x\to 0}\frac {x^2}{1-\sec(x)}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{1-\sec^2(x)}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{-\tan^2(x)}$$

Answer 1

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 $.

Answer 2

$\lim \limits_{x\to 0} \frac{x^2}{1-\sec x}= \lim \limits_{x\to 0} \frac {x^2}{1-\sec x}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{\cos^2 x-1}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{-\sin^2(x)}$

Now if you know the limit of $\frac {\sin x}x$ the other terms are well behaved.

Answer 3

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 $1-\cos^{2}x=\sin^{2}x$.

Answer 4

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 $$ \lim_{x \to 0}(1-\frac{x^2}{2} +O(x^4))\lim_{x \to 0}\frac{x^2}{-\frac{x^2}{2}+O(x^4)}=1 \cdot (-2)=-2 $$