limits calculus?

Question

Evaluate the limit.
lim_(x->infinity) x^(7/x)

Evaluate the limit.
lim_(x->infinity)

(sqrt(x^2 + 10 x) - x)

Answer 1

a.
f = x^(7/x)
=> ln f = 7ln (x)/x
x->infinity
lim ( ln f) = lim 7ln(x)/x = 7 lim ln(x)/x
= 7 lim[ (ln(x))'/ (x)' ]
= 7 lim( (1/x)/1) = 7 lim (1/x) = 0
=> lim (f) = e^0 =1
b.
x-> + infinity
sqrt(x^2 + 10 x) - x = [(sqrt(x^2 + 10 x) - x)* (sqrt(x^2 + 10 x) + x)/( sqrt(x^2 + 10 x) + x)
=(x2+10x -x2)/( sqrt(x^2 + 10 x) + x)
= 10x/( sqrt(x^2 + 10 x) + x)
= 10/( sqrt(1 + 10/x) + 1)
=> lim( sqrt(x^2 + 10 x) - x) = lim( 10/( sqrt(1 + 10/x) + 1))
= 10/(1+1) = 5
x-> - infinity
lim( sqrt(x^2 + 10 x) - x) = + infinity

Answer 2

lny = 7/x*lnx ==>0 as x==> infinity so lim y =1 as the ln function is continuos
Multiplying by the conjugate
=(x^2+10x -x^2)/(sqrt(x^2+10x)+x)
As x==> + infinity
= 10x/( x( sqrt(1+10/x)+1) and simplifying ===> 5