Evaluate the following limits as x approaches 0: (iv) lim (tan 2x / tan 5x); (v) lim (sin 3x / sin 7x); (vi) lim (sin(ax) / (bx)); (vii) lim (sin 4x / sin 2x); (viii) lim (x sec x)... Evaluate the following limits as x approaches 0: (iv) lim (tan 2x / tan 5x); (v) lim (sin 3x / sin 7x); (vi) lim (sin(ax) / (bx)); (vii) lim (sin 4x / sin 2x); (viii) lim (x sec x); (ix) lim (cos x); (x) lim ((sin(ax) + bx) / (ax + sin(bx))); where a, b, a + b ≠ 0. Read more

Steps to Solve

1. Evaluating the first limit: $\lim_{x \to 0} \frac{\tan 2x}{\tan 5x}$

Use the fact that $\tan x \approx x$ as $x$ approaches 0.

This means: $$ \lim_{x \to 0} \frac{\tan 2x}{\tan 5x} = \lim_{x \to 0} \frac{2x}{5x} = \frac{2}{5}. $$

2. Evaluating the second limit: $\lim_{x \to 0} \frac{\sin 3x}{\sin 7x}$

Using the approximation $\sin x \approx x$ as $x$ approaches 0, we have: $$ \lim_{x \to 0} \frac{\sin 3x}{\sin 7x} = \lim_{x \to 0} \frac{3x}{7x} = \frac{3}{7}. $$

3. Evaluating the third limit: $\lim_{x \to 0} \frac{\sin(ax)}{bx}$

Again, using $\sin x \approx x$: $$ \lim_{x \to 0} \frac{\sin(ax)}{bx} = \lim_{x \to 0} \frac{ax}{bx} = \frac{a}{b}. $$

4. Evaluating the fourth limit: $\lim_{x \to 0} x \sec x$

Using the fact that $\sec x = \frac{1}{\cos x}$, and $\cos x \approx 1$ near 0: $$ \lim_{x \to 0} x \sec x = \lim_{x \to 0} \frac{x}{\cos x} = \lim_{x \to 0} x = 0.$$

5. Evaluating the fifth limit: $\lim_{x \to 0} \frac{\sin 4x}{\sin 2x}$

Using the approximation for $\sin$: $$ \lim_{x \to 0} \frac{\sin 4x}{\sin 2x} = \lim_{x \to 0} \frac{4x}{2x} = 2. $$

6. Evaluating the sixth limit: $\lim_{x \to 0} \frac{\cos x}{\pi - x}$

Using the fact that $\cos x \approx 1$: $$ \lim_{x \to 0} \frac{\cos x}{\pi - x} = \frac{1}{\pi}. $$

7. Evaluating the seventh limit: $\lim_{x \to 0} \frac{\sin(ax + bx)}{ax + \sin(bx)}$

Apply small angle approximations: $$ \lim_{x \to 0} \frac{\sin((a+b)x)}{ax + (bx)} = \lim_{x \to 0} \frac{(a+b)x}{(a+b)x} = 1 \text{ (if } a, b \text{ are non-zero)}. $$