Problema 12

Calcular los siguientes límites si es que existen:

1. $\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3}$

$$\begin{align}\lim_{x\to 3}\frac{x^2-9}{x-3} &= \lim_{x\to 3}\frac{(x+3)(x-3)}{x-3}\\ &= \lim_{x\to 3}x+3\\&= 3+3=\boxed{6}\end{align}$$ 2. $\displaystyle\lim_{x\to -1}\frac{x+1}{x^3+1}$ $$\begin{align}\lim_{x\to -1}\frac{x+1}{x^3+1} &= \lim_{x\to -1}\frac{x+1}{(x+1)(x^2-x+1)}\\ &= \lim_{x\to -1}\frac{1}{x^2-x+1}\\&= \frac{1}{(-1)^2-(-1)+1}=\boxed{\frac{1}{3}}\end{align}$$ 3. $\displaystyle\lim_{x\to2}\frac{x-2}{x^2-4}$ $$\begin{align}\lim_{x\to2}\frac{x-2}{x^2-4} &= \lim_{x\to2}\frac{x-2}{(x+2)(x-2)}\\ &= \lim_{x\to2}\frac{1}{x+2}\\&= \frac{1}{2+2}=\boxed{\frac{1}{4}}\end{align}$$ 4. $\displaystyle\lim_{h\to0}\frac{(x+h)^2-x^2}{h}$ $$\begin{align}\lim_{h\to0}\frac{(x+h)^2-x^2}{h} &= \lim_{h\to0}\frac{x^2+2xh+h^2-x^2}{h}\\ &= \lim_{h\to0}\frac{2xh+h^2}{h}\\&= \lim_{h\to0} 2x+h\\&= 2x+0=\boxed{2x}\end{align}$$ 5. $\displaystyle\lim_{x\to-2}\frac{3x^2+2x-8}{x+2}$ ** Trabajo previo: $3x^2+2x-8=\frac{(3x)^2+2(3x)-24}{3}=\frac{(3x+6)(3x-4)}{3}=(x+2)(3x-4)$ $$\begin{align}\lim_{x\to-2}\frac{3x^2+2x-8}{x+2} &= \lim_{x\to-2}\frac{(x+2)(3x-4)}{x+2}\\ &= \lim_{x\to-2}3x-4\\&= 3(-2)-4\\&=\boxed{-10}\end{align}$$ 6. $\displaystyle\lim_{x\to2}\frac{\sqrt{2x-3}-\sqrt{x-1}}{x^2-4}$ $$\begin{align}\lim_{x\to2}\frac{\sqrt{2x-3}-\sqrt{x-1}}{x^2-4} &=\lim_{x\to2}\frac{\sqrt{2x-3}-\sqrt{x-1}}{x^2-4}\cdot\frac{\sqrt{2x-3}+\sqrt{x-1}}{\sqrt{2x-3}+\sqrt{x-1}}\\&=\lim_{x\to2}\frac{2x-3-x+1}{(x^2-4)(\sqrt{2x-3}+\sqrt{x-1})}\\ &= \lim_{x\to2}\frac{x-2}{(x+2)(x-2)(\sqrt{2x-3}+\sqrt{x-1})}\\ &=\lim_{x\to2}\frac{1}{(x+2)(\sqrt{2x-3}+\sqrt{x-1})}\\ &= \frac{1}{(2+2)(\sqrt{2(2)-3}+\sqrt{2-1})}\\&= \frac{1}{(4)(\sqrt{1}+\sqrt{1})}= \boxed{\frac{1}{8}}\end{align}$$ 7. $\displaystyle\lim_{x\to+\infty}\frac{4x^3+x^2-1}{5x^3+x}$ $$\begin{align}\lim_{x\to+\infty}\frac{4x^3+x^2-1}{5x^3+x} &= \lim_{x\to+\infty}\frac{\frac{4x^3}{x^3}+\frac{x^2}{x^3}-\frac{1}{x^3}}{\frac{5x^3}{x^3}+\frac{x}{x^3}}\\ &= \lim_{x\to+\infty}\frac{4+\frac{1}{x}-\frac{1}{x^3}}{5+\frac{1}{x^2}}\\&=\frac{4+\frac{1}{\infty}-\frac{1}{\infty^3}}{5+\frac{1}{\infty^2}}=\boxed{\frac{4}{5}}\end{align}$$ 8. $\displaystyle\lim_{x\to+\infty}\frac{(x+18)(x^4-x)}{x^5+1}$ $$\begin{align}\lim_{x\to+\infty}\frac{(x+18)(x^4-x)}{x^5+1} &= \lim_{x\to+\infty}\frac{x^5+18x^4-x^2-18x}{x^5+1}\\&= \lim_{x\to+\infty}\frac{\frac{x^5}{x^5}+\frac{18x^4}{x^5}-\frac{x^2}{x^5}-\frac{18x}{x^5}}{\frac{x^5}{x^5}+\frac{1}{x^5}}\\ &= \lim_{x\to+\infty}\frac{1+\frac{18}{x}-\frac{1}{x^3}-\frac{18}{x^4}}{1+\frac{1}{x^5}}\\&=\frac{1+\frac{18}{\infty}-\frac{1}{\infty^3}-\frac{18}{\infty^4}}{1+\frac{1}{\infty^5}}=\boxed{1}\end{align}$$ 9. $\displaystyle\lim_{x\to1}\frac{x+1}{x^3-1}$ $$\begin{align}\lim_{x\to1}\frac{x+1}{x^3-1} &= \frac{1+1}{1-1}\\ &= \frac{2}{0}\end{align}$$ El límite no existe. 10. $\displaystyle\lim_{x\to0}\frac{1-\cos(x)}{\sin(x)}$ $$\begin{align}\lim_{x\to0}\frac{1-\cos(x)}{\sin(x)} &= \lim_{x\to0}\frac{1-\cos(x)}{\sin(x)}\cdot\frac{1+\cos(x)}{1+\cos(x)}\\ &= \lim_{x\to0}\frac{1-\cos^2(x)}{\sin(x)(1+\cos(x))}\\ &=\lim_{x\to0}\frac{\sin^2(x)}{\sin(x)(1+\cos(x))}\\ &=\lim_{x\to0}\frac{\sin(x)}{1+\cos(x)}\\ &= \frac{\sin(0)}{1+\cos(0)}\\ &= \frac{0}{1+1}=\boxed{0}\end{align}$$ 11. $\displaystyle\lim_{x\to2}\frac{x^3-8}{x-2}$ $$\begin{align}\lim_{x\to2}\frac{x^3-8}{x-2} &= \lim_{x\to2}\frac{(x-2)(x^2+2x+4)}{x-2}\\ &= \lim_{x\to2} x^2+2x+4 \\&= 2^2+2(2)+4=\boxed{12}\end{align}$$ 12. $\displaystyle\lim_{x\to-\infty}\frac{3x^3-2x^2+7}{6x^4-x^3+2x-100}$ $$\begin{align}\lim_{x\to-\infty}\frac{3x^3-2x^2+7}{6x^4-x^3+2x-100} &= \lim_{x\to-\infty}\frac{\frac{3x^3}{x^4}-\frac{2x^2}{x^4}+\frac{7}{x^4}}{\frac{6x^4}{x^4}-\frac{x^3}{x^4}+\frac{2x}{x^4}-\frac{100}{x^4}}\\ &= \lim_{x\to-\infty}\frac{\frac{3}{x}-\frac{2}{x^2}+\frac{7}{x^4}}{6-\frac{1}{x}+\frac{2}{x^3}-\frac{100}{x^4}}\\&=\frac{\frac{3}{-\infty}-\frac{2}{(-\infty)^2}+\frac{7}{(-\infty)^4}}{6-\frac{1}{-\infty}+\frac{2}{(-\infty)^3}-\frac{100}{(-\infty)^4}}=\boxed{0}\end{align}$$ 13. $\displaystyle\lim_{x\to\infty}\frac{x^2-2x+5}{\sqrt{3x^4+2x+1}}$ $$\begin{align}\lim_{x\to+\infty}\frac{x^2-2x+5}{\sqrt{3x^4+2x+1}} &= \lim_{x\to+\infty}\frac{\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{5}{x^2}}{\sqrt{\frac{3x^4}{x^4}+\frac{2x}{x^4}+\frac{1}{x^4}}}\\ &= \lim_{x\to+\infty}\frac{1-\frac{2}{x}+\frac{5}{x^2}}{\sqrt{3+\frac{2}{x^3}+\frac{1}{x^4}}}\\ &= \frac{1-\frac{2}{\infty}+\frac{5}{\infty^2}}{\sqrt{3+\frac{2}{\infty^3}+\frac{1}{\infty^4}}}\ =\boxed{\frac{1}{\sqrt{3}}}\end{align}$$ 14. $\displaystyle\lim_{x\to y}\frac{x^n-y^n}{x-y}$ $$\begin{align}\lim_{x\to y}\frac{x^n-y^n}{x-y} &= \lim_{x\to y}\frac{(x-y)\left(\sum_{k=0}^{n-1}x^{n-k-1}y^k\right)}{x-y}\\ &= \lim_{x\to y}\sum_{k=0}^{n-1}x^{n-k-1}y^k\\ &= \sum_{k=0}^{n-1}y^{n-k-1}y^k = \sum_{k=0}^{n-1}y^{n-1} = \boxed{ny^{n-1}} \end{align}$$