Limit Proofs 8Ce606

1. **Problem:** Prove that $$\lim_{x \to 0} x^4 \cos\left(\frac{2}{x}\right) = 0$$. 2. **Formula and rules:** We use the Squeeze Theorem. Since $$-1 \leq \cos\left(\frac{2}{x}\right) \leq 1$$ for all $$x \neq 0$$, multiplying by $$x^4 \geq 0$$ gives: $$-x^4 \leq x^4 \cos\left(\frac{2}{x}\right) \leq x^4$$ 3. **Intermediate work:** As $$x \to 0$$, both $$-x^4$$ and $$x^4$$ approach $$0$$. 4. **Conclusion:** By the Squeeze Theorem, $$\lim_{x \to 0} x^4 \cos\left(\frac{2}{x}\right) = 0$$. --- 1. **Problem:** Prove that $$\lim_{x \to 0^+} \sqrt{x} e^{\sin\left(\frac{\pi}{x}\right)} = 0$$. 2. **Formula and rules:** Since $$-1 \leq \sin\left(\frac{\pi}{x}\right) \leq 1$$, we have $$e^{-1} \leq e^{\sin\left(\frac{\pi}{x}\right)} \leq e^{1}$$. 3. **Intermediate work:** Multiplying by $$\sqrt{x}$$ (which tends to 0 as $$x \to 0^+$$), we get $$e^{-1} \sqrt{x} \leq \sqrt{x} e^{\sin\left(\frac{\pi}{x}\right)} \leq e^{1} \sqrt{x}$$ 4. **Conclusion:** Both bounds tend to 0 as $$x \to 0^+$$, so by the Squeeze Theorem, $$\lim_{x \to 0^+} \sqrt{x} e^{\sin\left(\frac{\pi}{x}\right)} = 0$$. --- 1. **Problem:** Find $$\lim_{x \to 3} (2x + |x - 3|)$$. 2. **Intermediate work:** Consider left and right limits: - For $$x \to 3^-$$, $$|x-3| = 3 - x$$, so $$2x + |x-3| = 2x + (3 - x) = x + 3$$. At $$x=3$$, limit from left is $$3 + 3 = 6$$. - For $$x \to 3^+$$, $$|x-3| = x - 3$$, so $$2x + |x-3| = 2x + (x - 3) = 3x - 3$$. At $$x=3$$, limit from right is $$3(3) - 3 = 6$$. 3. **Conclusion:** Both one-sided limits equal 6, so $$\lim_{x \to 3} (2x + |x - 3|) = 6$$. --- 1. **Problem:** Find $$\lim_{x \to -6} \frac{2x + 12}{|x + 6|}$$. 2. **Intermediate work:** Note numerator: $$2x + 12 = 2(x + 6)$$. Rewrite limit as $$\lim_{x \to -6} \frac{2(x + 6)}{|x + 6|} = 2 \lim_{x \to -6} \frac{x + 6}{|x + 6|}$$. 3. **Consider one-sided limits:** - For $$x \to -6^-$$, $$x + 6 < 0$$, so $$\frac{x + 6}{|x + 6|} = \frac{\text{negative}}{\text{positive}} = -1$$. - For $$x \to -6^+$$, $$x + 6 > 0$$, so $$\frac{x + 6}{|x + 6|} = 1$$. 4. **Conclusion:** Left limit is $$2 \times (-1) = -2$$, right limit is $$2 \times 1 = 2$$. Since one-sided limits differ, the limit does not exist. --- 1. **Problem:** Find $$\lim_{x \to 0^-} \left(\frac{1}{x} - \frac{1}{|x|}\right)$$. 2. **Intermediate work:** For $$x \to 0^-$$, $$x < 0$$, so $$|x| = -x$$. Rewrite expression: $$\frac{1}{x} - \frac{1}{|x|} = \frac{1}{x} - \frac{1}{-x} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}$$. 3. **Limit:** As $$x \to 0^-$$, $$x$$ approaches 0 from negative side, so $$\frac{2}{x} \to -\infty$$. 4. **Conclusion:** The limit diverges to negative infinity; it does not exist as a finite number. --- 1. **Problem:** Find $$\lim_{x \to -2} \frac{2 - |x|}{2 + x}$$. 2. **Intermediate work:** Evaluate numerator and denominator near $$x = -2$$. - For $$x \to -2^-$$, $$x < -2$$, so $$|x| = -x$$ (since $$x$$ is negative, absolute value is positive). - For $$x \to -2^+$$, $$x > -2$$, still negative but closer to -2. Calculate numerator: $$2 - |x|$$ Calculate denominator: $$2 + x$$ 3. **Check numerator at $$x = -2$$:** $$| -2 | = 2$$, so numerator is $$2 - 2 = 0$$. Denominator at $$x = -2$$ is $$2 - 2 = 0$$. So we have an indeterminate form $$\frac{0}{0}$$. 4. **Rewrite numerator:** Since $$x$$ near $$-2$$ is negative, $$|x| = -x$$. So numerator: $$2 - |x| = 2 - (-x) = 2 + x$$. 5. **Simplify expression:** $$\frac{2 - |x|}{2 + x} = \frac{2 + x}{2 + x}$$. 6. **Cancel common factor:** $$\frac{\cancel{2 + x}}{\cancel{2 + x}} = 1$$ for $$x \neq -2$$. 7. **Conclusion:** $$\lim_{x \to -2} \frac{2 - |x|}{2 + x} = 1$$. --- **Final answers:** 31. $$0$$ 32. $$0$$ 33. $$6$$ 34. Does not exist 35. Does not exist (diverges to $$-\infty$$) 36. $$1$$