Question: $$\lim_{x\rightarrow 0^{+}} x \ \tan(\frac{\pi }{2}-x)$$
My Work:
To begin with I re-arranged the the question to the $\frac{\infty}{\infty}$ form:
$$\lim_{x\rightarrow 0^{+}} \frac{\tan(\frac{\pi }{2}-x)}{\frac{1}{x}}$$
Then applying L'Hospital's rule I got the following:
$$\lim_{x\rightarrow 0^{+}}\frac{\sec ^{2}(\frac{\pi }{2}-x)}{\frac{1}{x^{2}}}$$ which I rearranged to $$\lim_{x\rightarrow 0^{+}} \frac{x^{2}}{\cos ^{2}(\frac{\pi}{2}-x)}$$ Applying L'Hospital's rule again I got the following: $$\lim_{x\rightarrow 0^{+}} \frac{2x}{(-2)(-1))\cos(\frac{\pi }{2}-x)\sin(\frac{\pi }{2}-x) )} \ = \lim_{x\rightarrow 0^{+}} \frac{2x}{\sin(\pi-2x) )}$$
Above I used the property $\sin(2x) = 2\sin(x) \ \cos(x) $
And then applying L'Hospital's rule for the final time I got the following: $$ \lim_{x\rightarrow 0^{+}} \frac{-1}{\cos(\pi-2x) )} = 1$$
Is my answer and method correct?
Your method looks fine to me. However you need not apply L'Hospital's rule more than once because :
$$\lim_{x \to 0^+} \frac{x^2}{\cos^2(\pi/2 -x)} = \lim_{x \to 0^+}\frac{x^2}{\sin^2(x)} = \lim_{x \to 0^+}\frac{x^2}{\sin^2x} = \lim_{x \to 0^+} \left( \frac{x}{\sin x}\right)^2 = 1 $$
using $\cos(\pi/2 - x) = \sin(x)$ and the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$
As Jaap Scherphuis said in comments, we need not apply L Hospital's rule as $x \tan(\pi/2 - x) = \cos(x) \frac{x}{\sin(x)} \to 1$ as $x \to 0$ using the rules of limits.