limit as h approaches 0 of (sqrt(x^2+2xh+h^2+1))/h-(sqrt(x^2+1))/h

Lösung

h \to 0 lim (\frac{x ^{2} + 2 x h + h ^{2} + 1}{h} - \frac{x ^{2} + 1}{h})

\frac{x}{x ^{2} + 1}

Schritte zur Lösung

h \to 0 lim (\frac{x ^{2} + 2 x h + h ^{2} + 1}{h} - \frac{x ^{2} + 1}{h})

Multipliziere mit der Konjugation von

\frac{x ^{2} + 2 x h + h ^{2} + 1}{h} - \frac{x ^{2} + 1}{h} : \frac{\frac{2 x}{h} + 1}{\frac{x ^{2} + 2 x h + h ^{2} + 1}{h} + \frac{x ^{2} + 1}{h}}

= h \to 0 lim (\frac{\frac{2 x}{h} + 1}{\frac{x ^{2} + 2 x h + h ^{2} + 1}{h} + \frac{x ^{2} + 1}{h}})

Vereinfache

\frac{\frac{2 x}{h} + 1}{\frac{x ^{2} + 2 x h + h ^{2} + 1}{h} + \frac{x ^{2} + 1}{h}} : \frac{2 x + h}{x ^{2} + 2 x h + h ^{2} + 1 + x ^{2} + 1}

= h \to 0 lim (\frac{2 x + h}{x ^{2} + 2 x h + h ^{2} + 1 + x ^{2} + 1})

Setze den Wert

h = 0

ein

= \frac{2 x + 0}{x ^{2} + 2 x \cdot 0 + 0 ^{2} + 1 + x ^{2} + 1}

Vereinfache

\frac{2 x + 0}{x ^{2} + 2 x \cdot 0 + 0 ^{2} + 1 + x ^{2} + 1} : \frac{x}{x ^{2} + 1}

= \frac{x}{x ^{2} + 1}

t \to 0 lim (sin (\frac{1}{t}))

x \to \infty lim (x + z)

x \to 0 lim (\frac{sin ( x ) ( cos ( x ) - 1 )}{x ^{2}})

x \to \infty lim (x^{3} - 4 x^{2} + 2)

x \to 0 lim (\frac{13 e ^{2 x} - 13}{e ^{x} - 1})