Calculate the following limit:

Answers

Answer 1

Answer: $\lim_(x\to\infty)\frac{\sqrt x}{\sqrt{x+√(x+\sqrt x)}}=\n\lim_(x\to\infty)\frac{(\sqrt x)/(\sqrt x)}{\frac{\sqrt{x+√(x+\sqrt x)}}{\sqrt x}}=\n\lim_(x\to\infty)\frac{1}{\sqrt{(x+√(x+\sqrt x))/(x)}}=\n\lim_(x\to\infty)\frac{1}{\sqrt{1+(√(x+\sqrt x))/(x)}}=\n\lim_(x\to\infty)\frac{1}{\sqrt{1+(√(x+\sqrt x))/(√(x^2))}}=\n$
$\lim_(x\to\infty)\frac{1}{\sqrt{1+\sqrt{(x+\sqrt x)/(x^2)}}}=\n\lim_(x\to\infty)\frac{1}{\sqrt{1+\sqrt{(1)/(x)+(\sqrt x)/(√(x^4))}}}=\n\lim_(x\to\infty)\frac{1}{\sqrt{1+\sqrt{(1)/(x)+\sqrt{(x)/(x^4)}}}}=\n\lim_(x\to\infty)\frac{1}{\sqrt{1+\sqrt{(1)/(x)+\sqrt{(1)/(x^3)}}}}=\n=\frac{1}{\sqrt{1+\sqrt{0+√(0)}}}=\n$
$=(1)/(√(1+0))=\n=(1)/(√(1))=\n=(1)/(1)=\n1$

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Y is direcly proportional to x and y=27 when x=6find x when y=45

Answers

Answer:

x= 10

Step-by-step explanation:

27÷6= 4.5

27×4.5= 6

so,

45÷4.5=10

Answer:

7.5

Step-by-step explanation:

45 divided by 6 i don't know but I just wanted to try

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