find limit and use l'hospital's rule if appropriate.?

Question

1)find: as lim x-->infinity xtan(1/x)

2)find: as lim x-->infinity (e^x+x)^1/x

Answer 1

1) lim x tan(1/x) as x--> inf
= lim (tan (1/x))/ (1/x) as x --> inf
= lim (-1/x^2)(sec (1/x))^2 / (-1/x^2) as x--> inf
= lim (sec(1/x))^2 as x--> inf
= (sec 0)^2
= 1/(cos 0)^2
=1

2) lim (e^x+x)^(1/x) as x --> inf
= lim (e^x + e^(ln x))^(1/x) as x--> inf since x = e^(ln x)
= lim (e^(x+ ln x))^ (1/x) as x--> inf
= lim e^(1+((ln x)/x)) as x--> inf
= lim e.e^((ln x)/x) as x--> inf
= lim e.(e^(ln x))^(1/x) as x--> inf
= lim e.(x)^(1/x) as x--> inf
=e.(infinity)^(0) =e(1)
=e or 2.71828...
since every number (including infinity) to the power of zero is equal to1.

Answer 2

1) Find: as lim x→∞ xtan(1/x).

lim x→∞ xtan(1/x) = lim x→∞ tan(1/x) / (1/x)

Now we have the indeterminant form 0/0 and can apply L'Hospital's Rule.

lim x→∞ tan(1/x) / (1/x) = lim x→∞ (-1/x²)sec²(1/x) / (-1/x²)

= lim x→∞ sec²(1/x) = sec²(0) = 1
___________________________

2) Find: as lim x→∞ (e^x+x)^(1/x).

y = lim x→∞ (e^x+x)^(1/x)
ln y = lim x→∞ ln[(e^x+x)^(1/x)] = lim x→∞ ln(e^x+x) / x

Now we have the indeterminant form ∞/∞ and can apply L'Hospital's Rule.

ln y = lim x→∞ ln(e^x+x) / x = lim x→∞ [(e^x+1)/(e^x+x)] / 1

ln y = lim x→∞ [(e^x+1)/(e^x+x)]

ln y = lim x→∞ [e^x / (e^x+1)]

ln y = lim x→∞ e^x / e^x = 1

y = e^1 = e

Answer 3

lim x-->infinty tan(1/x)
let infinity=@ then
limx-->@ tan(1/x)
this will give us infinty after applying limit because
tan(1/@) =tan(0) and tan(0) =@
there for we apply L-hospital rule and we have to differentiate tan 1/x w.r.t x and then we will apply limit same is the case in the another one

Answer 4

1) 1/x *x* sin(1/x) / cos(1/x)
=sin(1/x)/(1/x) * 1/cos(1/x)

lim x-->infinity 1*1/cos(1/infinity)
=1/cos(0)= 1/1=1