First, let's evaluate the first limit:
lim x → 0 x + 1 + x + 4 − 3 x \lim_{x\to 0} \frac{\sqrt{x+1} + \sqrt{x+4} - 3}{x} lim x → 0 x x + 1 + x + 4 − 3 . We can rewrite the expression as:
lim x → 0 ( x + 1 − 1 ) + ( x + 4 − 2 ) x \lim_{x\to 0} \frac{(\sqrt{x+1}-1) + (\sqrt{x+4}-2)}{x} lim x → 0 x ( x + 1 − 1 ) + ( x + 4 − 2 ) . Now, we multiply by the conjugates:
lim x → 0 ( x + 1 − 1 ) ( x + 1 + 1 ) x ( x + 1 + 1 ) + ( x + 4 − 2 ) ( x + 4 + 2 ) x ( x + 4 + 2 ) \lim_{x\to 0} \frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{x(\sqrt{x+1}+1)} + \frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)} lim x → 0 x ( x + 1 + 1 ) ( x + 1 − 1 ) ( x + 1 + 1 ) + x ( x + 4 + 2 ) ( x + 4 − 2 ) ( x + 4 + 2 ) = lim x → 0 x + 1 − 1 x ( x + 1 + 1 ) + x + 4 − 4 x ( x + 4 + 2 ) = \lim_{x\to 0} \frac{x+1-1}{x(\sqrt{x+1}+1)} + \frac{x+4-4}{x(\sqrt{x+4}+2)} = lim x → 0 x ( x + 1 + 1 ) x + 1 − 1 + x ( x + 4 + 2 ) x + 4 − 4 = lim x → 0 x x ( x + 1 + 1 ) + x x ( x + 4 + 2 ) = \lim_{x\to 0} \frac{x}{x(\sqrt{x+1}+1)} + \frac{x}{x(\sqrt{x+4}+2)} = lim x → 0 x ( x + 1 + 1 ) x + x ( x + 4 + 2 ) x = lim x → 0 1 x + 1 + 1 + 1 x + 4 + 2 = \lim_{x\to 0} \frac{1}{\sqrt{x+1}+1} + \frac{1}{\sqrt{x+4}+2} = lim x → 0 x + 1 + 1 1 + x + 4 + 2 1 = 1 0 + 1 + 1 + 1 0 + 4 + 2 = 1 1 + 1 + 1 2 + 2 = 1 2 + 1 4 = 2 4 + 1 4 = 3 4 = \frac{1}{\sqrt{0+1}+1} + \frac{1}{\sqrt{0+4}+2} = \frac{1}{1+1} + \frac{1}{2+2} = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} = 0 + 1 + 1 1 + 0 + 4 + 2 1 = 1 + 1 1 + 2 + 2 1 = 2 1 + 4 1 = 4 2 + 4 1 = 4 3 .
Next, let's evaluate the second limit:
lim x → π 2 ( x tan x − π 2 cos x ) \lim_{x\to \frac{\pi}{2}} \left(x \tan x - \frac{\pi}{2\cos x}\right) lim x → 2 π ( x tan x − 2 c o s x π ) = lim x → π 2 ( x sin x cos x − π 2 cos x ) = \lim_{x\to \frac{\pi}{2}} \left(\frac{x \sin x}{\cos x} - \frac{\pi}{2\cos x}\right) = lim x → 2 π ( c o s x x s i n x − 2 c o s x π ) = lim x → π 2 2 x sin x − π 2 cos x = \lim_{x\to \frac{\pi}{2}} \frac{2x \sin x - \pi}{2\cos x} = lim x → 2 π 2 c o s x 2 x s i n x − π . Let x = π 2 + h x = \frac{\pi}{2} + h x = 2 π + h . As x → π 2 x \to \frac{\pi}{2} x → 2 π , h → 0 h \to 0 h → 0 . = lim h → 0 2 ( π 2 + h ) sin ( π 2 + h ) − π 2 cos ( π 2 + h ) = \lim_{h\to 0} \frac{2(\frac{\pi}{2}+h) \sin(\frac{\pi}{2}+h) - \pi}{2\cos(\frac{\pi}{2}+h)} = lim h → 0 2 c o s ( 2 π + h ) 2 ( 2 π + h ) s i n ( 2 π + h ) − π = lim h → 0 2 ( π 2 + h ) cos ( h ) − π − 2 sin ( h ) = \lim_{h\to 0} \frac{2(\frac{\pi}{2}+h) \cos(h) - \pi}{-2\sin(h)} = lim h → 0 − 2 s i n ( h ) 2 ( 2 π + h ) c o s ( h ) − π = lim h → 0 π cos h + 2 h cos h − π − 2 sin h = \lim_{h\to 0} \frac{\pi\cos h + 2h\cos h - \pi}{-2\sin h} = lim h → 0 − 2 s i n h π c o s h + 2 h c o s h − π = lim h → 0 π ( cos h − 1 ) + 2 h cos h − 2 sin h = \lim_{h\to 0} \frac{\pi(\cos h - 1) + 2h\cos h}{-2\sin h} = lim h → 0 − 2 s i n h π ( c o s h − 1 ) + 2 h c o s h = lim h → 0 π ( cos h − 1 ) − 2 sin h + lim h → 0 2 h cos h − 2 sin h = \lim_{h\to 0} \frac{\pi(\cos h - 1)}{-2\sin h} + \lim_{h\to 0} \frac{2h\cos h}{-2\sin h} = lim h → 0 − 2 s i n h π ( c o s h − 1 ) + lim h → 0 − 2 s i n h 2 h c o s h = lim h → 0 π ( cos h − 1 ) − 2 sin h − lim h → 0 h cos h sin h = \lim_{h\to 0} \frac{\pi(\cos h - 1)}{-2\sin h} - \lim_{h\to 0} \frac{h\cos h}{\sin h} = lim h → 0 − 2 s i n h π ( c o s h − 1 ) − lim h → 0 s i n h h c o s h = lim h → 0 π ( cos h − 1 ) − 2 sin h − lim h → 0 h sin h cos h = \lim_{h\to 0} \frac{\pi(\cos h - 1)}{-2\sin h} - \lim_{h\to 0} \frac{h}{\sin h}\cos h = lim h → 0 − 2 s i n h π ( c o s h − 1 ) − lim h → 0 s i n h h cos h Using L'Hopital's rule on the first term:
lim h → 0 π ( − sin h ) − 2 cos h = π ( 0 ) − 2 ( 1 ) = 0 \lim_{h\to 0} \frac{\pi(-\sin h)}{-2\cos h} = \frac{\pi(0)}{-2(1)} = 0 lim h → 0 − 2 c o s h π ( − s i n h ) = − 2 ( 1 ) π ( 0 ) = 0 . The second term is:
− lim h → 0 h sin h ⋅ lim h → 0 cos h = − 1 ⋅ 1 = − 1 - \lim_{h\to 0} \frac{h}{\sin h} \cdot \lim_{h\to 0} \cos h = -1 \cdot 1 = -1 − lim h → 0 s i n h h ⋅ lim h → 0 cos h = − 1 ⋅ 1 = − 1 . Therefore, the limit is 0 − 1 = − 1 0 - 1 = -1 0 − 1 = − 1 . 