Limit of a function without L'Hopital's rule.

Question

Find $$\lim _{x\to 0^+}\left(\frac{1}{x}-\sqrt{\frac{1}{x^2}-\frac{3}{x}}\right)$$

This was my quiz question (We can't use L'Hopital's rule as it is not covered yet.)

I know that if we divide a number by a significantly small number we will get a big number. Solving the question from there I got $+ \infty$

Did I do the question correctly? If not, can you show me the right way to do it?

Answer 1

The limit is not $+\infty$, and if it were, then it wouldn't be for the reasons you suggest; more has to be done.

This limit has an indeterminate form, so you're better off doing some algebraic manipulations first. Noting that $x>0$ (since the limit approaches $0$ from above), you can put the expression inside the square root over a common denominator of $x^2$ and pull that out of the root, to obtain $$\lim_{x \to 0^+} \dfrac{1- \sqrt{1-3x}}{x}$$ This still has an indeterminate form, since the top and bottom both tend to $0$ as $x \to 0^+$. But now you can apply l'Hôpital's rule to find the limit.

Edit: If you can't use L'Hôpital's rule, then an algebraic solution can be obtained by multiplying the numerator and denominator of the fraction by $1+\sqrt{1-3x}$.

Answer 2

write $$\frac{(1-\sqrt{1-3x})(1+\sqrt{1+3x})}{x(1+\sqrt{1-3x})}$$ the searched Limit is $$\frac{3}{2}$$ without L'Hospital