Answer:
[tex]\lim_{x\rightarrow +\infty } \frac{\sqrt{x^{2}+1} }{x+1}=1[/tex]
Step-by-step explanation:
[tex]\lim_{x\rightarrow +\infty } \frac{\sqrt{x^{2}+1} }{x+1}[/tex]
[tex]=\lim_{x\rightarrow +\infty } \frac{\sqrt{x^{2}\left( 1+\frac{1}{x^2} \right) } }{x+1}[/tex]
[tex]=\lim_{x\rightarrow +\infty } \frac{\sqrt{x^{2}} \sqrt{\left( 1+\frac{1}{x^2} \right) } }{x+1}[/tex]
[tex]=\lim_{x\rightarrow +\infty } \frac{|x|\sqrt{\left( 1+\frac{1}{x^2} \right) } }{x+1}[/tex]
[tex]=\lim_{x\rightarrow +\infty } \frac{x\sqrt{\left( 1+\frac{1}{x^2} \right) } }{x(1+\frac{1}{x}) }[/tex]
[tex]=\lim_{x\rightarrow +\infty } \frac{\sqrt{\left( 1+\frac{1}{x^2} \right) } }{(1+\frac{1}{x}) }[/tex]
[tex]=\frac{\sqrt{1} }{1}[/tex]
[tex]=1[/tex]
Remark :
[tex]\lim_{x\rightarrow +\infty } \frac{1}{x} =0=\lim_{x\rightarrow +\infty } \frac{1}{x^2}[/tex]
