Berechne die folgenden (eventuell uneigentlichen) Grenzwerte.

1. lim_(x -> pi/4) (sin(x) + cos(x))/(16x^2 + 1)
2. lim_(x -> oo) (x+1)^x/x^x
3. lim_(x -> 1) (x^3 - 6x^2 + 11x - 6)/(x^2 - 5x + 4)
Approach

lim_(x -> pi/4) (sin(x) + cos(x))/(16x^2 + 1)

Take the limit: lim_(x->pi/4) (cos(x)+sin(x))/(1+16 x^2)

The limit of a quotient is the quotient of the limits: = (lim_(x->pi/4) (cos(x)+sin(x)))/(lim_(x->pi/4) (1+16 x^2))

The limit of 1+16 x^2 as x approaches pi/4 is 1+pi^2: = (lim_(x->pi/4) (cos(x)+sin(x)))/(1+pi^2)

The limit of a sum is the sum of the limits: = (lim_(x->pi/4) cos(x)+lim_(x->pi/4) sin(x))/(1+pi^2)

The limit of cos(x) as x approaches pi/4 is 1/sqrt(2): = (1/sqrt(2)+lim_(x->pi/4) sin(x))/(1+pi^2)

The limit of sin(x) as x approaches pi/4 is 1/sqrt(2): = sqrt(2)/(1+pi^2)

lim_(x -> oo) (x+1)^x/x^x

Take the limit: lim_(x->infinity) (1+x)^x/x^x

Simplify (1+x)^x/x^x assuming x>0 giving (1+1/x)^x: = lim_(x->oo) (1+1/x)^x

Indeterminate form of type 1^oo. Transform using lim_(x->o) (1+1/x)^x = e^(lim_(x->oo) x log(1+1/x)): = e^(lim_(x->oo) x log(1+1/x))

Indeterminate form of type 0·oo. Let t = 1/x, then lim_(x->oo) x log(1+1/x) = lim_(t->0) (log(1+t))/t: = e^(lim_(t->0) (log(1+t))/t)

Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(t->0) (log(1+t))/t = lim_(t->0) (( dlog(1+t))/( dt))/(( dt)/( dt)): = e^(lim_(t->0) 1/(1+t))

The limit of a quotient is the quotient of the limits: The limit of a constant is the constant: = e^(1/(lim_(t->0) (1+t))) The limit of 1+t as t approaches 0 is 1:

= e

lim_(x -> 1) (x^3 - 6x^2 + 11x - 6)/(x^2 - 5x + 4)

Take the limit: lim_(x->1) (-6+11 x-6 x^2+x^3)/(4-5 x+x^2)

Factor the numerator and denominator: = lim_(x->1) ((-1+x) (6-5 x+x^2))/((-1+x) (-4+x))

Cancel terms, assuming -1+x!=0: = lim_(x->1) (6-5 x+x^2)/(-4+x)

The limit of a quotient is the quotient of the limits: = (lim_(x->1) (6-5 x+x^2))/(lim_(x->1) (-4+x))

The limit of -4+x as x approaches 1 is -3: = -1/3 (lim_(x->1) (6-5 x+x^2))

The limit of 6-5 x+x^2 as x approaches 1 is 2: = -2/3

Solution
1. sqrt(2)/(1+pi^2)
2. e
3. -2/3
• URL:
• Language:
• Subjects: math
• Type: Calculate
• Duration: 35min
• Credits: 6
• Difficulty: 0.5
• Tags: hpi limit function
• Note:
HPI, 2014-05-05, Mathe 2, Aufgabe 19
• Created By: adius
• Created At:
2014-07-26 10:17:45 UTC