-1
$\begingroup$

I need to solve this limit without L'Hôpital's rule. These questions always seem to have some algebraic trick which I just can't see this time...

$$\lim_{x \rightarrow 0}\left(\frac{\sqrt{7x+11}−\sqrt{11}}{5^{sin{\left(\sqrt{2x+3}−\sqrt{3}\right)}}−1}\right)$$

I motified it.

$\endgroup$
  • $\begingroup$ Welcome to the Math Stack Exchange. Did you intend the 3 in the denominator to be a square root of 3? $\endgroup$ – Paul Sundheim Oct 17 '16 at 14:43
  • $\begingroup$ When you change the problem, an apology is usually in order because members may have spent time on the incorrectly stated problem. $\endgroup$ – zhw. Oct 17 '16 at 16:15
1
$\begingroup$

The numerator & denominator are continuous so we see that the top is $0$ and the bottom is $5 \sin (\sqrt{3}-3) -1 < 0$, hence the limit is $0$.

Sorry, I should elaborate, note that $0 < 3- \sqrt{3} < 3 < \pi$, hence $\sin (\sqrt{3}-3) < 0$ and so $5 \sin (\sqrt{3}-3) -1 < 0$.

$\endgroup$
0
$\begingroup$

Hint for the newly stated problem: Denote the numerator by $f(x),$ the denominator by $g(x).$ The expression then equals

$$ \frac{f(x)-f(0)}{g(x)-g(0)} = \frac{(f(x)-f(0))/(x-0)}{(g(x)-g(0))(x-0)}.$$

As $x\to 0,$this $\to f'(0)/g'(0)$ by the definition of the derivative. (We are not using L'Hopital.)

$\endgroup$
  • $\begingroup$ is method different to L'Hopital? $\endgroup$ – user35446 Oct 17 '16 at 17:24
  • $\begingroup$ Yes it is $\,\,\,$ $\endgroup$ – zhw. Oct 17 '16 at 17:48
  • $\begingroup$ Thanks! I got a same value that determine by using L'Hopital. $\endgroup$ – user35446 Oct 17 '16 at 18:11

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.