Spart Posted March 20, 2013 Share Posted March 20, 2013 Hello, I'm having a bit of difficulty calculating this limit. [latex]\lim_{x\to+\infty }\frac{1}{\; \sqrt{x+1}-\sqrt{x+7}\; }[/latex] [latex]\lim_{x\to+\infty }\frac{1}{\; \sqrt{+\infty +1}-\sqrt{+\infty +7}\; }[/latex] [latex]\lim_{x\to+\infty }\frac{1}{\; \sqrt{+\infty}-\sqrt{+\infty}\; }[/latex] [latex]\lim_{x\to+\infty }\frac{1}{\;+\infty-\infty \; }[/latex] Well this is where I'm stuck. Naturally the square root of infinity will always remain infinity however it results in an indeterminate form since positive infinity cannot be summed with negative infinity. According to my text book the answer should be [latex]-\infty[/latex] but I have no idea how to work it out. Link to comment Share on other sites More sharing options...
Yash Posted March 20, 2013 Share Posted March 20, 2013 try rationalisation 1 Link to comment Share on other sites More sharing options...
Spart Posted March 20, 2013 Author Share Posted March 20, 2013 try rationalisation [latex]\lim_{x\to+\infty }\frac{1}{\; \sqrt{x+1}-\sqrt{x+7}\; } \times \frac{\sqrt{x+1}+\sqrt{x+7}}{\sqrt{x+1}+\sqrt{x+7}}[/latex] [latex]\lim_{x\to+\infty }\frac{\sqrt{x+1}+\sqrt{x+7}}{\left ( \sqrt{x+1}-\sqrt{x+7} \right )\left ( \sqrt{x+1}+\sqrt{x+7} \right )}[/latex] [latex]\lim_{x\to+\infty }\frac{\sqrt{x+1}+\sqrt{x+7}}{\sqrt{x+1}^{\: 2}-\sqrt{x+7}^{\: 2}}[/latex] [latex]\lim_{x\to+\infty }\frac{\sqrt{x+1}+\sqrt{x+7}}{\left ( x+1 \right )-\left ( x+7 \right )}[/latex] [latex]\lim_{x\to+\infty }\frac{\sqrt{x+1}+\sqrt{x+7}}{-6} =\frac{+\infty +\infty }{-6}=-\infty[/latex] Simple! Thank you! Link to comment Share on other sites More sharing options...
Yash Posted March 21, 2013 Share Posted March 21, 2013 just remember that limits are generally applied in the last step. Link to comment Share on other sites More sharing options...
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