1. **Problem Statement:** Explain why the expression $1^\infty$ is an indeterminate form in calculus, using the example $$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$$ and provide another example where the limit is actually 1. 2. **Definition:** An indeterminate form arises when the limit of an expression cannot be directly determined from the limits of its parts. The form $1^\infty$ is indeterminate because the base approaches 1 while the exponent grows without bound, leading to competing effects. 3. **Example 1:** Consider $$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$$ - Using the definition of $e$, this limit equals $e$ (approximately 2.718), not 1. - This shows that even though the base tends to 1, the exponent tends to infinity, and the overall limit is not 1. 4. **Example 2:** Consider $$\lim_{x \to 0} 1^\frac{1}{x}$$ - Here, the base is exactly 1 for all $x$, so the limit is 1. - The difference is that the base is constant 1, not approaching 1. 5. **Summary:** The indeterminate form $1^\infty$ requires careful limit evaluation because the base and exponent's behavior can produce different limits. --- 6. **Problem Statement:** Define indeterminate form and evaluate the following limits: (a) $$\lim_{x \to 0} \left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}$$ (b) $$\lim_{x \to 0} \left(\frac{1 + \tan x}{1 + \sin x}\right)^{\frac{1}{x^2}}$$ (c) $$\lim_{x \to 0^+} \left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$ (d) $$\lim_{x \to 0} \frac{x e^{-1 + x^2} \log(1 + x)}{x^2}$$ (e) $$\lim_{x \to +\infty} \left(\frac{\sqrt{x^2 + \sqrt{x^4 + 25}}}{3}\right)^x$$ 7. **Definition:** An indeterminate form is a limit expression where substitution leads to ambiguous forms like $0/0$, $\infty/\infty$, $1^\infty$, etc., requiring further analysis. 8. **Solution (a):** - Use series expansions: $\tan x \approx x + \frac{x^3}{3}$, so $\frac{\tan x}{x} \approx 1 + \frac{x^2}{3}$. - Then $$\left(1 + \frac{x^2}{3}\right)^{\frac{1}{x^2}} = e^{\frac{1}{x^2} \ln\left(1 + \frac{x^2}{3}\right)} \approx e^{\frac{1}{x^2} \cdot \frac{x^2}{3}} = e^{\frac{1}{3}}.$$ - **Answer:** $e^{1/3}$. 9. **Solution (b):** - Approximate numerator and denominator: $1 + \tan x \approx 1 + x + \frac{x^3}{3}$, $1 + \sin x \approx 1 + x - \frac{x^3}{6}$. - Ratio $$\approx \frac{1 + x + \frac{x^3}{3}}{1 + x - \frac{x^3}{6}} = 1 + \frac{x^3/3 + x^3/6}{1 + x - x^3/6} \approx 1 + \frac{x^3}{2}.$$ - Then $$\left(1 + \frac{x^3}{2}\right)^{\frac{1}{x^2}} = e^{\frac{1}{x^2} \ln\left(1 + \frac{x^3}{2}\right)} \approx e^{\frac{1}{x^2} \cdot \frac{x^3}{2}} = e^{\frac{x}{2}} \to 1.$$ - **Answer:** 1. 10. **Solution (c):** - $\frac{\sin x}{x} \approx 1 - \frac{x^2}{6}$. - Then $$\left(1 - \frac{x^2}{6}\right)^{\frac{1}{x}} = e^{\frac{1}{x} \ln\left(1 - \frac{x^2}{6}\right)} \approx e^{\frac{1}{x} \cdot \left(-\frac{x^2}{6}\right)} = e^{-\frac{x}{6}} \to 1.$$ - **Answer:** 1. 11. **Solution (d):** - Simplify numerator: $x e^{-1 + x^2} \log(1 + x) \approx x e^{-1} (x - \frac{x^2}{2}) = x e^{-1} x = x^2 e^{-1}$ ignoring higher order terms. - Then $$\frac{x e^{-1 + x^2} \log(1 + x)}{x^2} \approx \frac{x^2 e^{-1}}{x^2} = e^{-1}.$$ - **Answer:** $e^{-1}$. 12. **Solution (e):** - For large $x$, $$\sqrt{x^2 + \sqrt{x^4 + 25}} \approx \sqrt{x^2 + x^2} = \sqrt{2} x.$$ - So expression inside limit is $$\left(\frac{\sqrt{2} x}{3}\right)^x = \left(\frac{\sqrt{2}}{3} x\right)^x.$$ - Since $x \to +\infty$, and $\frac{\sqrt{2}}{3} x \to \infty$, the limit is infinite. - **Answer:** The limit diverges to infinity. --- 13. **Problem Statement:** Find $a$ such that $$\lim_{x \to 0} \frac{a \sin x - \sin 2x}{\tan^3 x}$$ is finite. 14. **Solution:** - Use expansions: $\sin x \approx x - \frac{x^3}{6}$, $\sin 2x \approx 2x - \frac{8x^3}{6} = 2x - \frac{4x^3}{3}$, $\tan x \approx x + \frac{x^3}{3}$ so $\tan^3 x \approx x^3$. - Numerator $$a \sin x - \sin 2x \approx a \left(x - \frac{x^3}{6}\right) - \left(2x - \frac{4x^3}{3}\right) = (a - 2) x + \left(-\frac{a}{6} + \frac{4}{3}\right) x^3.$$ - Denominator $$\tan^3 x \approx x^3.$$ - Limit becomes $$\lim_{x \to 0} \frac{(a - 2) x + \left(-\frac{a}{6} + \frac{4}{3}\right) x^3}{x^3} = \lim_{x \to 0} \frac{(a - 2)}{x^2} + \left(-\frac{a}{6} + \frac{4}{3}\right).$$ - For limit to be finite, coefficient of $\frac{1}{x^2}$ must be zero: $$a - 2 = 0 \Rightarrow a = 2.$$ - Then limit is $$-\frac{2}{6} + \frac{4}{3} = -\frac{1}{3} + \frac{4}{3} = 1.$$ - **Answer:** $a = 2$, limit = 1. --- 15. **Problem Statement:** Find $a,b,c$ such that $$\lim_{x \to 0} \frac{a + b \cos 3x - c \sin x}{x^2}$$ is finite. 16. **Solution:** - Use expansions: $\cos 3x \approx 1 - \frac{9x^2}{2}$, $\sin x \approx x$. - Numerator $$a + b \left(1 - \frac{9x^2}{2}\right) - c x = (a + b) - c x - \frac{9b}{2} x^2.$$ - Divide by $x^2$: $$\frac{a + b - c x - \frac{9b}{2} x^2}{x^2} = \frac{a + b}{x^2} - \frac{c}{x} - \frac{9b}{2}.$$ - For limit to be finite, coefficients of $\frac{1}{x^2}$ and $\frac{1}{x}$ must be zero: $$a + b = 0,$$ $$c = 0.$$ - Limit then is $$-\frac{9b}{2}.$$ - **Answer:** $c=0$, $a = -b$, limit finite and equals $-\frac{9b}{2}$.