Find the limit

[Leon234556]

Hello. Someone help me with this limit, please. (I'm new in the forum)

 

[Dr.Peterson]

Hello. Someone help me with this limit, please. (I'm new in the forum)

What have you tried? What methods have you learned? Where do you need help?

I'd first try multiplying and dividing by the conjugate. There will be a further trick or two needed!

Show us as far as you get, by that method or another.

 

[topsquark]

[math]\lim_{x \to - \infty} \dfrac{\sqrt{x^2 + x + 1} + x}{1}[/math]
Hint: Start by rationalizing the numerator.

-Dan

 

[Deleted member 4993]

[math]\lim_{x \to - \infty} \dfrac{\sqrt{x^2 + x + 1} + x}{1}[/math]
Hint: Start by rationalizing the numerator.

-Dan

Hint: use

a^2 - b^2 = (a+b) * (a-b)

What have you tried? What methods have you learned? Where do you need help?

 

[lex]

Here is a similar example:

 

[nasi112]

I would prefer this method. You need to be familiar with L'hopital Rule.

[MATH]\lim_{x\to -\infty} \sqrt{x^2 + x + 1} + x = \lim_{x\to \infty} \sqrt{x^2 - x + 1} - x = \lim_{x\to \infty} x\left(\frac{\sqrt{x^2 - x + 1} - x}{x}\right)[/MATH]

[MATH]= \lim_{x\to \infty} x\left(\sqrt{\frac{x^2}{x^2} - \frac{x}{x^2} + \frac{1}{x^2}} - \frac{x}{x}\right) = \lim_{x\to \infty} x\left(\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} - 1\right)[/MATH]

[MATH]= \lim_{x\to \infty} \frac{\left(\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} - 1\right)}{\frac{1}{x}} = \frac{0}{0}[/MATH]
Now, you can use L'hopital Rule, which means you can differentiate the numerator and the denominator seperately.

[MATH]= \lim_{x\to \infty} \frac{\left(\frac{\frac{1}{x^2} - \frac{2}{x^3}}{2\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}}\right)}{\frac{-1}{x^2}} = \lim_{x\to \infty} \frac{\left(\frac{-1}{x^2}\right)\left(-1 + \frac{2}{x}\right)}{\left(\frac{-1}{x^2}\right)2\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}} \ = \ ?[/MATH]

 

[lex]

The original suggested solution avoids using a 'big' theorem and avoids an unpleasant differentiation.
Yours is a well-worked solution, however!

Spoiler

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]

 

[nasi112]

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 another beautiful solution.

 

[lex]

A slight variation.

Spoiler

[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).

 

[nasi112]

A slight variation.

Spoiler

[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).

This is also beautiful, but I always prefer to convert [MATH]-\infty[/MATH] to [MATH]\infty[/MATH].

It eliminates a lot of possible calculation mistakes.

 

[lookagain]

Hello. Someone help me with this limit, please. (I'm new in the forum)

\(\displaystyle \sqrt{a^2 + b} \ \approx \ a+ \tfrac{b}{2a} \ \) for b-values that are relatively small in comparison
to the a-value, and a, b > 0.

You can change it to the limit as x approaches +oo of \(\displaystyle \sqrt{x^2 - x + 1} - x.\)

Here, \(\displaystyle \ a^2 = x^2 \ \ and \ \ b = -x + 1 \)

 

[lookagain]

[MATH]= \lim_{x\to \infty} \frac{\left(\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} - 1\right)}{\frac{1}{x}} = [/MATH]

\(\displaystyle \lim_{x\to 0^+} \dfrac{\sqrt{1 - x + x^2 \ } - 1}{x}\)

nasi112, it would make it easier to change it to this form before using L'hopital's Rule.