# Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$

Evaluate $$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$

I assumed $x=\frac{1}{y}$ we get

$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$

using L'Hopital's rule we get

$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$

$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$

is this possible to do without Lhopita's rule

• Actually you just applied the definition of derivative. – egreg Jul 21 '17 at 17:41

## 8 Answers

Well, we can start as you, by setting $y=\frac{1}{x}$. Now, our limits transforms to: $$L=\lim_{y\to0}\frac{\tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}}{y}$$ Now, let $f:\mathbb{R}\to\mathbb{R}$ with $$f(x)=\tan\left(\frac{1+x}{1+4x}\right)$$ Note that $f(0)=\tan^{-1}(1)=\frac{\pi}{4}$. So, we have: $$L=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)$$ since $f$ is differentiable. Now, $f$'s formula is: $$f(x)=\int_0^\frac{1+x}{1+4x}\frac{1}{1+t^2}dt$$ So, we have: $$f'(x)=\frac{1}{1+\left(\frac{1+x}{1+4x}\right)^2}\left(\frac{1+x}{1+4x}\right)'=\frac{(1+4x)^2}{1+(1+x)^2}\frac{-3}{(1+4x)^2}=\frac{-3(1+4x)^2}{(1+(1+x)^2)(1+4x)^2}$$ So $$L=f'(0)=-\frac{3}{2}$$ As you have already proved.

Yes Possible.

Evaluate

$$\lim_{ x\to \infty} \left( tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$

I assumed $x=\frac{1}{y}$ we get

$$L=\lim_{y \to 0}\frac{\left( tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$

$$L=\lim_{y \to 0}\frac{\left( tan^{-1}\left(\frac{1+y}{1+4y}\right)-\tan^{-1}{1}\right)}{y}$$

using the following formula, we can acheive, $$tan^{-1}A-tan^{-1}B = \frac{A-B}{1+AB}$$ we get

$$L=\lim_{y \to 0} \frac{1}{y} \times \frac{-3y}{(2+5y)}$$

$$L=\lim_{y \to 0} \frac{-3}{(2+5y)}=\frac{-3}{2}$$

HINT:

Set $1/x=h$ to get $$F=\lim_{h\to0^+}\dfrac{\tan^{-1}\dfrac{h+1}{4h+1}-\tan^{-1}1}h$$

Now $\tan^{-1}\dfrac{h+1}{4h+1}-\tan^{-1}1=\tan^{-1}\left(\dfrac{\dfrac{h+1}{4h+1}-1}{1+\dfrac{h+1}{4h+1}}\right)=\tan^{-1}\dfrac{-3h}{5h+2}$

$$\implies F=\lim_{h\to0^+}\dfrac{\tan^{-1}\left(\dfrac{-3h}{5h+2}\right)}{\left(\dfrac{-3h}{5h+2}\right)}\cdot\lim_{h\to0^+}\dfrac{-3}{5h+2}$$

Let $\arctan\frac{1+x}{4+x}=\frac{\pi}{4}+y$.

Hence, $y\rightarrow0$ and $$\frac{1+x}{4+x}=\tan\left(\frac{\pi}{4}+y\right)$$ or $$x=\frac{1-4\tan\left(\frac{\pi}{4}+y\right)}{\tan\left(\frac{\pi}{4}+y\right)-1}$$ and we need to calculate $$\lim_{y\rightarrow0}\frac{y\left(1-4\tan\left(\frac{\pi}{4}+y\right)\right)}{\tan\left(\frac{\pi}{4}+y\right)-1}$$ or $$-\frac{3}{\sqrt2}\lim_{y\rightarrow0}\frac{y}{\sin\left(\frac{\pi}{4}+y\right)-\cos\left(\frac{\pi}{4}+y\right)}$$ or $$-\frac{3}{\sqrt2}\lim_{y\rightarrow0}\frac{y}{\frac{2}{\sqrt2}\sin{y}},$$ which is $-\frac{3}{2}$.

It would be much simpler to proceed directly without any change of variables. Note that the expression under limit can be written as $$x\arctan\dfrac{\dfrac{1+x}{4+x}-1}{\dfrac{1+x}{4+x}+1}$$ which is same as $$-\frac{3x}{5+2x}\cdot\dfrac{\arctan\dfrac{3}{5+2x}}{\dfrac{3}{5+2x}}$$ and first factor tends to $-3/2$ and second factor tends to $1$ so that the desired limit is $-3/2$.

Let $$y=\arctan\left(\frac{1+x}{4+x}\right).$$ Then $$x=\frac{1}{2}\left[3\left(\frac{1+\tan y}{1-\tan y}\right)-5\right].$$ As $x \to \infty, y \to \frac{\pi}{4}.$ So your limit is \begin{align*} \lim_{ x\to \infty} \left( \arctan\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x & = \lim_{y \to \pi/4}\left(y-\frac{\pi}{4}\right)\frac{1}{2}\left[3\left(\frac{1+\tan y}{1-\tan y}\right)-5\right]\\ & = \lim_{y \to \pi/4}\left(y-\frac{\pi}{4}\right)\frac{1}{2}\left[3\left(\frac{-1}{\tan\left(y-\frac{\pi}{4}\right)}\right)-5\right] \end{align*} Now with $\theta=y-\pi/4$, use the limit $\lim_{\theta \to 0}\frac{\tan \theta}{\theta}=1$ to get the answer as $-3/2$.

Using the formula for the tangent of a sum, $$\tan\left(x-\frac\pi4\right)=\frac{\tan(x)-1}{\tan(x)+1}$$ we get $$\tan^{-1}(x)-\frac\pi4=\tan^{-1}\left(\frac{x-1}{x+1}\right)$$ Thus, \begin{align} \lim_{x\to\infty}\left(\tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac\pi4\right)x &=\lim_{x\to\infty}\tan^{-1}\left(\frac{-3}{5+2x}\right)x\\ &=\lim_{x\to\infty}\frac{\tan^{-1}\left(\frac{-3}{5+2x}\right)}{\frac{-3}{5+2x}}\lim_{x\to\infty}\frac{-3x}{5+2x}\\ &=-\frac32 \end{align}

\begin{aligned} \lim _{x\to \infty }\left(\arctan\left(\frac{1+x}{4+x}\right)-\frac{\pi \:}{4}\right)x & = \lim _{t\to 0}\left(\frac{\arctan \left(\frac{t+1}{4t+1}\right)-\frac{\pi }{4}}{t}\right) \\ & = \lim _{t\to 0}\left(\frac{\left(\frac{\pi \:}{4}-\frac{3}{2}t+o\left(t\right)\right)-\frac{\pi \:}{4}}{t}\right)\\ & = \color{red}{-\frac{3}{2}} \end{aligned}

Solved with Taylor expansion: $$\arctan \left(\frac{t+1}{4t+1}\right) = \frac{\pi }{4}-\frac{3t}{2}+\frac{15t^2}{4}-\frac{33t^3}{4}+15t^4+\ldots \:$$