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.

try rationalisation

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!

just remember that limits are generally applied in the last step.