0
$\begingroup$

Show that $(\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}})$ tends to a limit as $x \rightarrow 0$, and evaluate the limit.

Taking the L'Hopital's rule, I get $$\frac{\frac{2x+1}{\sqrt{1+x+x^2}}}{(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1-x}})}$$

So as $x \rightarrow 0$, the equation tends $1/2$.

Is this correct?

$\endgroup$
3
$\begingroup$

The problem is that $\sqrt{1+x+x^2}$ is not zero when x tends to zero, it means that you have no an undetermined term of the form $0/0$ to justify the application of the L'Hopital theorem. You should replace at first and check that the expression is suitable to apply the L'Hopital rule if not you will get a wrong calculation of the limit.

$\endgroup$
2
  • $\begingroup$ sorry I forgot to add a -1 to the original problem $\endgroup$
    – spruce
    Mar 24 '20 at 17:34
  • 1
    $\begingroup$ In this case, your answer $1/2$ is right and your procedure is right also. $\endgroup$
    – Camilo160
    Mar 24 '20 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.