I Need Help With Limits!?

Question

Can you help me find the limits of the following problems? Explanations are appreciated too. Thanks!

1. lim x-> 0+ [sqrt(x^2 + 4x + 5) - sqrt(5)] / h

2. lim x-> 2+ (x + 3) times [(absolute value (x + 2)) / (x + 2)]

3. lim x-> 2- (x + 3) times [(absolute value (x + 2)) / (x + 2)]

Answer 1

The first one is 0 if for sure, the deno is h and not x.

for 2 and 3 make cases as x increases to 2 [x tends to 2 and x < 2]and x decreases to 2.

If x<2 the x+2 < 0 and |x+2| = - (x+2) = -x-2. so for second problem,

it is (x+3)(- 1) etc.

Answer 2

1. lim x-> 0+ [sqrt(x^2 + 4x + 5) - sqrt(5)] /x=

Multiply both the numerator and the denominator by [√(x^2+4x+5)+√5]

lim (x^2+4x+5-5)/[x(√(x^2+4x+5)+√5]=

x->0⁺

lim x(x+4)/[x(√(x^2+4x+5)+√5]=

x->0⁺

lim (x+4)/[(√(x^2+4x+5)+√5]=4/(√5+√5)=

x->0⁺

2/√5=(2√5)/5

2. lim x->2⁺ (x+3)*|(x+2)|/(x+2)=

lim  (x+3)*(x+2)/(x+2)=

x->2⁺

lim (x+3)*1=2+3=5

x->2⁺

3. lim x->2⁻ (x+3)*|(x+2)|/(x+2)=

lim  (x+3)*(x+2)/(x+2)=

x->2⁻

lim (x+3)*1=2+3=5

x->2⁻

⋆⋆ BUT

a) lim x->-2⁺ (x+3)*|(x+2)|/(x+2)=

lim  (x+3)*(x+2)/(x+2)=

x->-2⁺

lim [(x+3)*1]=-2+3=1

x->-2⁺

b) lim x->-2⁻ (x+3)*|(x+2)|/(x+2)=

lim  (x+3)*[-(x+2)]/(x+2)=

x->-2⁻

lim [(x+3)*(-1)]=-(-2+3)=-1

x->-2⁻

Because x+2>0 for x>-2 and x+2<0 for x<-2.

Or (in other words) (x+2) changes its sigh as x passes through (-2),

but (x+2) doesn't change its sigh as x passes through (+2).