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Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$

My attempt:

Rationalizing:

$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$

$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$

$$=\lim_{x\to -\infty}\frac{7x}{\sqrt {4x^2+7x}-2x}$$

Dividing numerator and denominator by x:

$$=\lim_{x\to -\infty} \frac{7}{\sqrt{4+\frac{7}{x}}-2}$$

$$= \frac{7}{\sqrt{4+\frac{7}{-\infty}}-2}$$

$$= \frac{7}{\sqrt{4+0}-2}$$

$$=\frac{7}{2-2}$$

$$=\infty$$

Conclusion: Limit does not exist.

Why is my solution wrong?

Correct answer: $\frac{-7}{4}$

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In these cases, in order to avoid mistakes with the sign, we can let $y=-x\to \infty$ to obtain

$$\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}=\lim_{y\to \infty} \frac{-7y}{\sqrt {4y^2-7y}+2y}=-\frac 74$$

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Because, when you divide the denominator by $x$, you forgot $x$ is supposed to be negative, so that $$x=-\sqrt{x^2}.$$

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  • $\begingroup$ Sir, but x is already negative. And division of x by x should yield a positive value. $\endgroup$ – Soumee Aug 22 '18 at 14:43
  • $\begingroup$ Division of $x$ by $x$is no problem.The problem is when you make it enter the radical: $\frac1x\sqrt{4x^2+7x}=\color{red}-\sqrt{4+\frac7x}$. $\endgroup$ – Bernard Aug 22 '18 at 14:47
  • $\begingroup$ Got it. But why do we have to take $-x$ out explicitly when $x$ is under radical sign? Why is it not implicit? $\endgroup$ – Soumee Aug 22 '18 at 14:53
  • $\begingroup$ I don't see your point here. I just replaced $\frac1x$ with $-\sqrt{\frac1{x^2}}$. $\endgroup$ – Bernard Aug 22 '18 at 14:59
  • $\begingroup$ Understood somewhat. Will refer to texts. Thank you, sir. $\endgroup$ – Soumee Aug 22 '18 at 15:03
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hint When $x$ goes to $-\infty,$ it becomes negative .

on the other hand,

we have $$\boxed{\sqrt{x^2}=|x|}$$

the mistake you made can be corrected by $$\sqrt{(-x)^2}=-x$$.

In the denominator, factor out by $(-x)^2$.

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Hint: It is $$\sqrt{4x^2+7x}=\sqrt{x^2\left(4+\frac{7}{x}\right)}=-x\sqrt{4+\frac{7}{x}}$$

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It's a very common error: $\sqrt{x^2}=-x$, when $x<0$.

I usually suggest the substitution $t=-1/x$, so the limit becomes $$ \lim_{t\to0^+}\left(\sqrt{\frac{4}{t^2}-\frac{7}{t}}-\frac{2}{t}\right) = \lim_{t\to0^+}\frac{\sqrt{4-7t}-2}{t} $$ which is an easy derivative: $$ f(t)=\sqrt{4-7t} \qquad f'(t)=\frac{-7}{2\sqrt{4-7t}} \qquad f'(0)=-\frac{7}{4} $$ or, with a Taylor expansion, $$ \lim_{t\to0^+}\frac{\sqrt{4-7t}-2}{t} = \lim_{t\to0^+}\frac{2(\sqrt{1-7t/4}-1)}{t} = \lim_{t\to0^+}\frac{2\bigl(1-\frac{1}{2}\frac{7t}{4}+o(t)\bigr)}{t}=-\frac{7}{4} $$

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