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.

  • $\begingroup$ 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

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 $5 \sin (\sqrt{3}-3) -1 < 0$.


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.)

  • $\begingroup$ 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

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