 Hi and how are you all? Today my name is Priyanka and let us discuss this question. It says evaluate limit x approaches infinity under root x square plus x plus 1 minus x. Let us proceed with the solution. Let us rewrite the given limit once again given limit x approaches infinity bracket under root x square plus x plus 1 minus x bracket close. In the next step let us rationalize the numerator. We can do that by multiplying the given limit by under root x square plus x plus 1 plus x upon under root x square plus x plus 1 plus x. This implies we have in the numerator now x square plus x plus 1 minus x square and in the denominator we have under the root x square plus x plus 1 on simplifying the numerator we have limit x approaches infinity x plus 1 upon under root x square plus x plus 1 plus on splitting the numerator we approaches infinity x upon under root x square plus x plus 1 approaches infinity 1 upon dividing the numerator and denominator by x we have divided by under root we will divide it by x that means x square will come inside so x square upon x square plus x upon x square plus 1 upon x square plus x upon infinity 1 upon x square plus x plus 1 in square root solving we have it as 1 this also has one. So limit x approaches infinity 1 upon this will also get divided 1 upon under root 1 plus 1 upon x plus 1 upon x square plus 1 plus limit x approaches infinity 1 upon under root x square as x plus 1 plus x, further on using the limits now we are left with 1 upon under root 1 plus 1 plus 0 which further implies 1 upon 2 plus 0 which is further equal to 1 upon 2. So this is the required answer to the given question right. So hope you understood the whole concept well do take care of your calculations and have a very nice day ahead.