Answered

At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Calculate the following limit:

Calculate The Following Limit class=

Sagot :

konrad509
[tex]\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\ \lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\ \lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\ \lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\ \lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\[/tex]
[tex]\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\ \lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\ =\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\[/tex]
[tex]=\dfrac{1}{\sqrt{1+0}}=\\ =\dfrac{1}{\sqrt{1}}=\\ =\dfrac{1}{1}=\\ 1 [/tex]

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.