Question

help me please with the following limite

Answer 1

because infinitive is greater than all the number u can write it as ( 2x/2x)^x
it's equal to 1^x and the answer is 1 .

Answer 2

According to mathematica the answer is e (2.72)

Answer 3

this is the graph

Answer 4

dear in infinitive it approach 1 as i learnt .

Answer 5

wow very comple, but even in the graph y=1 as x tends to infinite. Im gonna choose your answer

Answer 6

thanks dear

Answer 7

this is what i got\[\lim_{x \rightarrow \infty}(\frac{2x+3}{2x+1})^{x+1}=\]\[\lim_{x \rightarrow \infty}(1-(\frac{2}{2x+3})^{x+1})\]\[=1-\lim_{x \rightarrow \infty} (\frac{2}{2x+3})^x \times \frac{2}{2x+3}\]but\[\lim_{x \rightarrow \infty}\frac{2}{2x+3}=0\]so =1-0=1

Answer 8

thanks Sarah

Answer 9

@sarah :
i think u made a mistake .
i think it's ( 1- (2/2x+2))^(x+1)

Answer 10

i mean you can take 1 out of bracket and then power the rest with x+1