Evaluate the  limits:SOLUTION : by substituting the limit directly we are going to obtain zero(0) which shows that the limit does not exist. 




Now to proof that the limits exists we simply rationalize the function since function is a Rational Function Involving a root at the denominator. 








By expanding the bracket we obtain. 









 factoring out 2+√x Yields








Simply take the limits  and that's it. 


















To evaluate this limit, we use the sandwich or squeeze Theorem of functions which implies that for every composite functions involving a trigonometric function then like -1≤f(x) g(x) ≤1 where f(x)=x and g(x)=sin1/x, 
Now multiply -1≤f(x) g(x) ≤1 by f(x)=x we get -x≤xsin1/x≤x
And taking the limits Yields 









Taking the limits of this function directly Yields 1,  and we can also take the limit by factorizing the denominator as shown below








And that's the answer, hence the limit exists at x=0. 








Taking The limit of this function directly, simply substitute the function and you get a zero(0) which means the function does not exist at x=1.










We use another approach to show the existence of the limits. We are going to use analytical approach by using left hand side and right. hand side limits. This simply Yields 0.5 or 1/2.