Find $\lim_{h \to 0}\frac{\tan\sqrt{x+h}-\tan\sqrt{x}}{\log(1+3h)}$ without L'Hospital's rule

Question

What is the limit of $$\lim_{h \to 0}\frac{\tan\sqrt{x+h}-\tan\sqrt{x}}{\log(1+3h)}$$

I am confused with square root function and L'Hospital's rule is prohibited . The professor requires a detailed solution. I don't know what to do. Thanks!

Answer 1

By the difference quotient definition of derivative, $$f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$$

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So dividing and multiplying your function with $h$, we get two derivatives, $1$ in numerator, and $2$ in denominator:

1. $$\lim_{h\to 0}\frac{\tan\sqrt{x+h}-\tan\sqrt{x}}{h} = \dfrac{d}{dx}(\tan\sqrt{x})$$

2. $$\lim_{h\to 0}\frac{\log(1+3h)-\log1}{3h}\times 3= 3\dfrac{d}{dx}(\log x)$$

Dividing $1$ by $2$ gives the limit:

$$\lim_{h \to 0}\frac{\tan\sqrt{x+h}-\tan\sqrt{x}}{\log(1+3h)}=\dfrac{\dfrac{d}{dx}(\tan\sqrt{x})}{3\dfrac{d}{dx}(\log x)}=\frac{\sec^2(\sqrt{x})}{6\sqrt{x}}$$

Answer 2

Hint: divide by $h$ both the top and bottom.

Answer 3

$$ \frac{\tan\sqrt{x+h}-\tan\sqrt{x}}{\log(1+3h)} = \frac{\left( \dfrac{\tan\sqrt{x+h}-\tan\sqrt{x}} h \right)}{\left( 3\cdot \dfrac{\log(1+3h) - \log 1} {3h} \right)}\ \longrightarrow\ \frac{\dfrac d {dx} \tan\sqrt x}{3 \left.\dfrac d {dx} \log x \right\vert_{x=1} } \text{ as } h\to0. $$

Answer 4

$$\tan(\sqrt{x+h}-\sqrt x)=\dfrac{\tan(\sqrt{x+h})-\tan\sqrt x}{1+\tan(\sqrt{x+h})\tan\sqrt x}$$

$$\dfrac{\tan(\sqrt{x+h})-\tan\sqrt x}{\ln(1+3h)}$$

$$=(1+\tan(\sqrt{x+h})\tan\sqrt x)\cdot\dfrac{\tan(\sqrt{x+h}-\sqrt x)}{\sqrt{x+h}-\sqrt x}\cdot\dfrac1{(\sqrt{x+h}+\sqrt x)}\cdot\dfrac{3h}{3\ln(1+3h)}$$

Now use

$\lim_{u\to0}\dfrac{\tan u}u=\lim_{u\to0}\dfrac{\sin u}u\cdot\dfrac1{\lim_{u\to0}\cos u}=1$

and $\lim_{v\to0}\dfrac{\ln(1+v)}v=1$

Answer 5

You can also use the trigonometric identity $$\tan(p)-\tan(q)=\frac{\sin(p-q)}{\cos(p)\cos(q)}$$ which gives $$\tan(\sqrt{x+h})-\tan(\sqrt{x})=\frac{\sin(\sqrt{x+h}-\sqrt{x})}{\cos(\sqrt{x+h})\cos(\sqrt{x})}$$ Now $$\sqrt{x+h}-\sqrt{x}=(\sqrt{x+h}-\sqrt{x})\times \frac{\sqrt{x+h}+\sqrt{x} }{\sqrt{x+h}+\sqrt{x} }=\frac h{\sqrt{x+h}+\sqrt{x} }$$ Since $h$ is small, then $$\sin(\sqrt{x+h}-\sqrt{x})=\sin\Big(\frac h{\sqrt{x+h}+\sqrt{x} }\Big)\simeq\frac h {2\sqrt x}$$ So, concerning the numerator $$\tan(\sqrt{x+h})-\tan(\sqrt{x})\simeq \frac h {2\sqrt x}\times \frac 1 {\cos^2(\sqrt x)}$$ Concerning the denominator $$\log(1+3h)\simeq 3h$$ Combining all together $$\frac{\tan\sqrt{x+h}-\tan\sqrt{x}}{\log(1+3h)}\simeq \frac h {2\sqrt x}\times \frac 1 {\cos^2(\sqrt x)}\times \frac 1 {3h}=\frac 1 {6\sqrt x}\times \frac 1 {\cos^2(\sqrt x)}=\frac{\sec^2(\sqrt{x})}{6\sqrt{x}}$$