Límites de funciones - problemas y soluciones

Solución:
[tex]\lim_{x\rightarrow 0} \frac{1}{x}\times \left(\frac{1}{x+4}-\frac{1}{4}\right)=\\=\lim_{x\rightarrow 0}\frac{4-4-x}{4x(x+4)}=\frac{-1}{4(x+4)}=\frac{-1}{16}[/tex]

Solución:
[tex]\lim_{x\rightarrow 4} \frac{1}{x-4}\times\left(\frac{1}{x}-\frac{1}{4}\right)=\\=\lim_{x\rightarrow 4} \frac{1}{x-4}\times\left(\frac{4-x}{4x}\right)=\lim_{x\rightarrow 4} \frac{-1}{4x}=\frac{-1}{16}[/tex]

Solución:
[tex]\lim_{x\rightarrow 2} \left(1-\frac{2}{x}\right)\times \left(\frac{3}{4-x^2}\right)= \lim_{x\rightarrow 2} \frac{-3(x-2)}{x(x-2)(x+2)}=\lim_{x\rightarrow 2}\frac{-3}{x(x+2)}=-\frac{3}{8}[/tex]

Solución:
[tex]\lim_{x\rightarrow -2} \frac{(x^2-x-6)^2}{(x+2)^2}=\lim_{x\rightarrow -2} \frac{(x+2)^2(x-3)^2}{(x+2)^2}=25[/tex]

Solución:
[tex]\lim_{x\rightarrow 5} \frac{\sqrt{x^2+5}-\sqrt{30}}{x-5}=\lim_{x\rightarrow 5} \frac{(\sqrt{x^2+5}-\sqrt{30})(\sqrt{x^2+5}+\sqrt{30})}{(x-5)(\sqrt{x^2+5}+\sqrt{30})}=\\=\lim_{x\rightarrow 5} \frac{x^2-25}{(x-5)(\sqrt{x^2+5}+\sqrt{30})}=\lim_{x\rightarrow 5} \frac{x+5}{\sqrt{x^2+5}+\sqrt{30}}=\frac{5}{\sqrt{30}}=\sqrt{\frac{5}{6}}[/tex]

Solución:
[tex]\lim_{x\rightarrow -3^-} \frac{x+3}{\sqrt{x^2-12x-45}}=\lim_{x\rightarrow -3^-} -\frac{x+3}{\sqrt{(x+3)(x-15)}}=\\=\lim_{x\rightarrow -3^-} \frac{-\sqrt{(x+3)^2}}{\sqrt{(x+3)(x-15)}}=-\lim_{x\rightarrow -3^-}\sqrt{\frac{(x+3)^2}{(x+3)(x-15)}}=-\lim_{x\rightarrow -3^-}\sqrt{\frac{x+3}{x-15}}=0[/tex]

Solución:
Note que:
[tex]\sqrt{x}-\sqrt{8}=(\sqrt[4]{x}-\sqrt[4]{8})(\sqrt[4]{x}+\sqrt[4]{8})[/tex]
Por lo tanto
[tex]\lim_{x\rightarrow 8} \frac{\sqrt[4]{x}-\sqrt[4]{8}}{\sqrt{x}-\sqrt{8}}=\lim_{x\rightarrow 8} \frac{\sqrt[4]{x}-\sqrt[4]{8}}{(\sqrt[4]{x}-\sqrt[4]{8})(\sqrt[4]{x}+\sqrt[4]{8})}=\\=\lim_{x\rightarrow 8} \frac{1}{\sqrt[4]{x}+\sqrt[4]{8}}=\frac{1}{2\sqrt[4]{8}}[/tex]

Solución:
[tex]\lim_{h\rightarrow 0} \frac{1-\frac{1}{h^2}}{1-\frac{1}{h}}=\lim_{h\rightarrow 0} \frac{\frac{h^2-1}{h^2}}{\frac{h-1}{h}}=\lim_{h\rightarrow 0} \frac{h+1}{h}=1 + \lim_{h\rightarrow 0} \frac{1}{h}[/tex] no existe porque cuando h->0^- el límite es -∞ y h->0⁺ el límite es +∞

Solución:
[tex]\lim_{x\rightarrow +1} \Rightarrow 1 < x < 2 \Rightarrow \frac{1}{2} < \frac{x}{2} < 1 \Rightarrow \begin{cases}0 < \frac{x}{2}-\frac{1}{2} < \frac{1}{2} \Rightarrow \lfloor\frac{x-1}{2}\rfloor=0 \\ 1 < \frac{x}{2}+\frac{1}{2} < \frac{3}{2} \Rightarrow \lfloor\frac{x+1}{2}\rfloor=1\end{cases} \Rightarrow \lim_{x\rightarrow +1}\left(\lfloor\frac{x-1}{2}\rfloor + \lfloor\frac{x+1}{2}\rfloor\right)=1\ \ \ (*)[/tex]
[tex]\lim_{x\rightarrow -1} \Rightarrow 0 < x < 1 \Rightarrow 0 < \frac{x}{2} < \frac{1}{2} \Rightarrow \begin{cases}-\frac{1}{2} < \frac{x}{2}-\frac{1}{2} < 0 \Rightarrow \lfloor\frac{x-1}{2}\rfloor=-1 \\ \frac{1}{2} < \frac{x}{2}+\frac{1}{2} < 1 \Rightarrow \lfloor\frac{x+1}{2}\rfloor=0\end{cases} \Rightarrow \lim_{x\rightarrow -1}\left(\lfloor\frac{x-1}{2}\rfloor + \lfloor\frac{x+1}{2}\rfloor\right)=-1\ \ \ (**)[/tex]
Por lo tanto de (*) y (**), [tex]\lim_{x\rightarrow 1} \left( \lfloor\frac{x-1}{2}\rfloor + \lfloor\frac{x+1}{2}\rfloor \right)[/tex] - no existe.

Solución:
[tex]\lim_{x\rightarrow 2} \frac{\sqrt{x^2+5}-3}{x^2-2x}=\lim_{x\rightarrow 2} \frac{(\sqrt{x^2+5}-3)(\sqrt{x^2+5}+3)}{(x^2-2x)(\sqrt{x^2+5}+3)}=\\=\lim_{x\rightarrow 2}\frac{x^2-4}{x(x-2)(\sqrt{x^2+5}+3)}=\lim_{x\rightarrow 2}\frac{x+2}{x(\sqrt{x^2+5}+3)}=\frac{1}{3}[/tex]

Solución:
[tex]\lim_{h\rightarrow 1} \frac{\sqrt{b+2(h-1)}-\sqrt{b}}{h-1}=\lim_{h\rightarrow 1} \frac{(\sqrt{b+2(h-1)}-\sqrt{b})(\sqrt{b+2(h-1)}+\sqrt{b})}{(h-1)(\sqrt{b+2(h-1)}+\sqrt{b})}=\\=\lim_{h\rightarrow 1} \frac{2(h-1)}{(h-1)(\sqrt{b+2(h-1)}+\sqrt{b})}=\frac{1}{\sqrt{b}}[/tex]

Solución:
[tex]\lim_{x\rightarrow 0} \frac{x sen x}{\sqrt{x^2+1}-1}=\lim_{x\rightarrow 0} \frac{(x sen x)(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)}=\\=\lim_{x\rightarrow 0} \frac{(x sen x)(\sqrt{x^2+1}+1)}{x\times x}=1\times 1\times 2=2[/tex]

Solución:
Sea cos^-1 x = t por lo tanto x =cos t. Si [tex]x \rightarrow 1^{-}[/tex] entonces [tex]t \rightarrow 0[/tex]
[tex]\lim_{x \rightarrow 1^{-}}\frac{(\cos^{-1} x)^2}{1-x^2}=\lim_{t\rightarrow 0}\frac{t^2}{1-\cos^2 t}=\\=\lim_{t\rightarrow 0}\frac{t^2}{\text{sen}^2 t}=1[/tex]

Solución:
$x^2 - x - 2 = (x - 2)(x + 1)$
$x^3 - 12x + 16 = (x - 2)^2(x + 4)$

[tex]\lim_{x\rightarrow 2} \frac{(x^2-x-2)^{40}}{(x^3-12x+16)^{20}}=\lim_{x\rightarrow 2} \frac{(x-2)^{40}(x+1)^{20}}{(x-2)^{40}(x+4)^{20}}=\\=\frac{3^{40}}{6^{20}}=\frac{3^{20}}{2^{20}}=(\frac{3}{2})^{20}[/tex]

Solución:
Cuando [tex]x\rightarrow 2^-[/tex], |x-2|= -(x-2)
[tex]\frac{(x+1)^3-27}{|x-2|}=\frac{(x+1-3)((x+1)^2+3(x+1)+9)}{-|x-2|}= -((x+1)^2+3(x+1)+9)[/tex]
Por lo tanto
[tex]\lim_{x\rightarrow 2^{-}} -\frac{(x+1)^3-27}{|x-2|}= \lim_{x\rightarrow 2^{-}} -(x+1)^2+3(x+1)+9=-27[/tex]