Please help with all questions! Greatly appreciated :) lim_x rightarrow 0 f(x) i

Question

Please help with all questions! Greatly appreciated :)

lim_x rightarrow 0 f(x) if f(x) = {x^2 + 1 x is rational 1 x is irrational lim_x rightarrow 1+ (1/x - 1 - 1/|x - 1|) lim_x rightarrow 1- (1/x - 1 - 1/|x - 1|) lim_x rightarrow 0 (tan(x)/x) lim_x rightarrow 0 (sin(5x)/3x) lim_x rightarrow 0 (sin^2(7x)/x^2) lim_x rightarrow infinity (squareroot x^4 - 2x^2 + 4/x^2 + 3x + 8) lim_x rightarrow -infinity (squareroot x^4 - 2x^2 + 4/x^2 + 3x + 8) lim_x rightarrow infinity (squareroot x^2 + 1 - x) lim_x rightarrow -infinity (squareroot x^2 + 1 - x)

Explanation / Answer

d) lim x ---> 0   ( tan x / x)

applying l'hopital rule

lim x ---> 0   ( sec^2x / 1)

lim x ---> 0   ( 1 / cos^2 x )

plug x = 0

we get     1 / cos^2 (0) = 1

hence,

lim x ---> 0   ( tan x / x)   = 1

e) lim x --->0 ( sin 5x / 3x )

applying L'hopital's rule

lim x --->0   ( cos (5x) 5 ) / 3

plug x = 0

( cos (5*0) 5 ) / 3

= 5 / 3

hence ,

lim x --->0 ( sin 5x / 3x ) = 5/3

f) lim x ---> 0 ( sin^2 (7x) / x^2 )

( sin^2 (7x) / x^2 )   = (( sin 7x / x))^2

= lim x ---> 0 (( sin 7x / x))^2

applying L'hopitals rule

lim x ---> 0 (( cos 7x)*7 / 1))^2

plug x = 0

{cos ( 7*0)*7 / 1}^2

= 49

hence , lim x ---> 0 ( sin^2 (7x) / x^2 ) = 49

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