Leon234556
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- Jun 9, 2021
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What have you tried? What methods have you learned? Where do you need help?Hello. Someone help me with this limit, please. (I'm new in the forum)
Hint: use[math]\lim_{x \to - \infty} \dfrac{\sqrt{x^2 + x + 1} + x}{1}[/math]
Hint: Start by rationalizing the numerator.
-Dan
What have you tried? What methods have you learned? Where do you need help?
This is another beautiful solution.The original suggested solution avoids using a 'big' theorem and avoids an unpleasant differentiation.
Yours is a well-worked solution, however!
The original (and the best??):
[MATH]\lim \limits_{x \to -\infty}(\sqrt{x^2+x+1}+x)\\ \text {is } \hspace1ex \lim \limits_{x \to \infty}(\sqrt{x^2-x+1}-x)[/MATH](replacing x with -x)
Now we are ultimately dealing with positive values of x
[MATH](\sqrt{x^2-x+1}-x)=\dfrac{(\sqrt{x^2-x+1}-x)(\sqrt{x^2-x+1}+x)}{\sqrt{x^2-x+1}+x}\\ =\dfrac{(x^2-x+1-x^2)}{\sqrt{x^2-x+1}+x}\\ =\dfrac{-x+1}{\sqrt{x^2-x+1}+x}\\[/MATH]Dividing top and bottom by [MATH]x≠0[/MATH][MATH]=\dfrac{-1+\tfrac{1}{x}}{\sqrt{1-\tfrac{1}{x}+\tfrac{1}{x^2}}+1}\\ \text{Now } \hspace1ex \lim \limits_{x \to \infty} \dfrac{-1+\tfrac{1}{x}}{\sqrt{1-\tfrac{1}{x}+\tfrac{1}{x^2}}+1}\\ =?[/MATH]
This is also beautiful, but I always prefer to convert [MATH]-\infty[/MATH] to [MATH]\infty[/MATH].A slight variation.
[MATH]\lim \limits_{x \to -\infty}(\sqrt{x^2+x+1}+x)\\ \text{ }\\ (\sqrt{x^2+x+1}+x)=\dfrac{(\sqrt{x^2+x+1}+x)(\sqrt{x^2+x+1}-x)}{\sqrt{x^2+x+1}-x}\\ =\dfrac{(x^2+x+1-x^2)}{\sqrt{x^2+x+1}-x}\\ =\dfrac{x+1}{\sqrt{x^2+x+1}-x}\\[/MATH]Dividing top and bottom by [MATH]x≠0[/MATH][MATH]=\dfrac{1+\tfrac{1}{x}}{-\sqrt{1+\tfrac{1}{x}+\tfrac{1}{x^2}}-1}\\ \text{Now } \hspace1ex \lim \limits_{x \to -\infty} \dfrac{1+\tfrac{1}{x}}{-\sqrt{1+\tfrac{1}{x}+\tfrac{1}{x^2}}-1}\\ =?[/MATH]
[MATH]\text{Note: since [MATH]x[/MATH] is ultimately negative in this limit, then }\hspace1ex \dfrac{1}{x} \sqrt{a} \hspace1ex \text{ is negative } \hspace1ex =\boldsymbol{-} \sqrt{\dfrac{a}{x^2}} [/MATH]
([MATH]a[/MATH] is ultimately positive in this limit).
Hello. Someone help me with this limit, please. (I'm new in the forum)
[MATH]= \lim_{x\to \infty} \frac{\left(\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} - 1\right)}{\frac{1}{x}} = [/MATH]