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Is there a way to solve such limit without using L'Hopital rule? $$\lim_{h \to 0} \frac{2\sin(\frac{1}{2}\ln(1+\frac{h}{x}))}{h} $$ Result should be $\frac{1}{x}$.

Edit: Thank you for your help, everything got really simple knowing, that $\frac{\ln(1+x)}{x} \to 1$ when $x \to 0$.

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    Have you tried anything so far? A series expansion might help... – 2017-02-07
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    Use limits like $$\lim_{x\to 0}\dfrac{\sin x} {x} =\lim_{x\to 0}\dfrac{\log(1+x)}{x}=1$$ and obtain the result as $1/x$. – 2017-02-07

3 Answers 3

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Let $f(t) = 2 \sin ((1/2)\ln(1+t)).$ The expression equals

$$\frac{1}{x}\frac{f(h/x)-f(0)}{h/x-0}.$$

By definition of the derivative, this expression $\to (1/x)f'(0)$ as $h\to 0.$ A simple computation shows $f'(0)=1,$ so the the limit is $1/x.$

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This quotient is a rate of change w.r.t. h: $$\frac{2\sin\Bigl(\frac{1}{2}\ln\bigl(1+\frac{h}{x}\bigr)\Bigr)}{h}=\frac{2\sin\Bigl(\frac{1}{2}\ln\bigl(1+\frac{h}{x}\bigr)\Bigr)-2\sin\Bigl(\frac{1}{2}\ln\bigl(1+\frac{0}{x}\bigr)\Bigr)}{h-0},$$ so its limit is the derivative at $h=0$: \begin{align}\biggl(2\sin\Bigl(\frac{1}{2}\ln\Bigl(1+\frac{h}{x}\Bigr)\Bigr)\biggr)'_{h=0}&= \left[2\cos\Bigl(\frac{1}{2}\ln\Bigl(1+\frac{h}{x}\Bigr)\Bigr)\frac1{2x\Bigl(1+\dfrac{h}{x}\Bigr)}\right]_{h=0}\\ &=\left[\frac{\cos\Bigl(\frac{1}{2}\ln\Bigl(1+\frac{h}{x}\Bigr)\Bigr)}{x+h}\right]_{h=0}=\frac1x. \end{align}

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$2\sin\left(\frac12ln(1+\frac hx)\right)$=$2\sin\left(\frac12\frac hx\frac{ln\left(1+\frac hx\right)}{\frac hx}\right)$

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    $\frac { 2sin(\frac { h }{ 2x } ) }{ h } ->\frac { 2 }{ 2x } \frac { sin(\frac { h }{ 2x } ) }{ \frac { h }{ 2x } } $ – 2017-02-07