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pasta
find the lim (x+1/x) as x tends to 0.and sketch a graph to support for answer.
\[lim_{x \rightarrow 0}(x+\frac{1}{x})=lim_{x \rightarrow 0}(\frac{x^2+1}{x})\] From here you can see that the function gets larger and larger as x approaches 0 from the left and the right.|dw:1349841452926:dw| also, if you look at \[lim_{x \rightarrow \infty}(\frac{x^2+1}{x})=2\] \[lim_{x \rightarrow -\infty}(\frac{x^2+1}{x})=2\]from l'Hopital's rule
as x tends to infinity the function is undefined as far as i know
then the x^2 cancels out.....the eq left is 1/x^2...... now when x approaches 0 the eq. approaches infinity,...!!1 so infinity is the ans
graph is this and its correct
range of the grph is (-infinity , -2] union [2,infinity)
@adi171 there is no way i can use l 'HOpitals rule becoz the function gives 1/0 not 0/0 OR infnty/infnty.
when i use the rule my derivative is 2x which gives the answer 0.HOW ARE YOU differentiating 2x/1 which is actualy the derivative .and it does not work for the rule on this case
you are finding the minimum and maximum points of this graph. as x tends to inifinity the function tends to infinity .and AS X tends to 0 the function is undefined(see the attached graph)