Limit Evaluation

Calculus Level 3

lim ⁡ x → 0 − 25 tan ⁡ − 1 a + 48 tan ⁡ − 1 ( a + x ) − 36 tan ⁡ − 1 ( a + 2 x ) + 16 tan ⁡ − 1 ( a + 3 x ) − 3 tan ⁡ − 1 ( a + 4 x ) 12 x = ? \small \lim_{x \to 0} \frac{-25 \tan^{-1}{a} + 48 \tan^{-1}(a+x) - 36 \tan^{-1}(a+2x) + 16 \tan^{-1}(a+3x) - 3 \tan^{-1}(a+4x)}{12x}=?

a ∈ R a \in \mathbb{R}

Bonus: Evaluate this limit without using L'Hôpital's rule . More importantly, how do you interpret this limit?

1 1 + a 2 \frac{1}{1+a^2} − 1 1 + a 2 -\frac{1}{1+a^2} a 2 1 + a 2 \frac{a^2}{1+a^2} − a 1 + a 2 -\frac{a}{1+a^2}

Karan Chatrath by Karan Chatrath , 27, Netherlands

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chris Lewis
Mar 4, 2021

We can rewrite the limit as L = lim ⁡ x → 0 [ 48 tan ⁡ − 1 ( a + x ) − 48 tan ⁡ − 1 a 12 x − 36 tan ⁡ − 1 ( a + 2 x ) − 36 tan ⁡ − 1 a 12 x + 16 tan ⁡ − 1 ( a + 3 x ) − 1 6 tan ⁡ − 1 a 12 x − 3 tan ⁡ − 1 ( a + 4 x ) − 3 tan ⁡ − 1 a 12 x ] = lim ⁡ x → 0 4 tan ⁡ − 1 ( a + x ) − tan ⁡ − 1 a x − lim ⁡ x → 0 6 tan ⁡ − 1 ( a + 2 x ) − tan ⁡ − 1 a 2 x + lim ⁡ x → 0 4 tan ⁡ − 1 ( a + 3 x ) − tan ⁡ − 1 a 3 x − lim ⁡ x → 0 tan ⁡ − 1 ( a + 4 x ) − tan ⁡ − 1 a 4 x = lim ⁡ x → 0 4 tan ⁡ − 1 ( a + x ) − tan ⁡ − 1 a x − lim ⁡ x 1 → 0 6 tan ⁡ − 1 ( a + x 1 ) − tan ⁡ − 1 a x 1 + lim ⁡ x 2 → 0 4 tan ⁡ − 1 ( a + x 2 ) − tan ⁡ − 1 a x 2 − lim ⁡ x 3 → 0 tan ⁡ − 1 ( a + x 3 ) − tan ⁡ − 1 a x 4 \begin{aligned} L&=\lim_{x \to 0} \left[ \frac{48\tan^{-1} (a+x)-48\tan^{-1} a}{12x}-\frac{36\tan^{-1} (a+2x)-36\tan^{-1} a}{12x}+\frac{16\tan^{-1} (a+3x){-1}6\tan^{-1} a}{12x}-\frac{3\tan^{-1} (a+4x)-3\tan^{-1} a}{12x}\right] \\ &=\lim_{x \to 0} 4\frac{\tan^{-1} (a+x)-\tan^{-1} a}{x}-\lim_{x \to 0}6\frac{\tan^{-1} (a+2x)-\tan^{-1} a}{2x}+\lim_{x \to 0} 4\frac{\tan^{-1} (a+3x)-\tan^{-1} a}{3x}-\lim_{x \to 0}\frac{\tan^{-1} (a+4x)-\tan^{-1} a}{4x} \\ &=\lim_{x \to 0} 4\frac{\tan^{-1} (a+x)-\tan^{-1} a}{x}-\lim_{x_1 \to 0}6\frac{\tan^{-1} (a+x_1)-\tan^{-1} a}{x_1}+\lim_{x_2 \to 0} 4\frac{\tan^{-1} (a+x_2)-\tan^{-1} a}{x_2}-\lim_{x_3 \to 0}\frac{\tan^{-1} (a+x_3)-\tan^{-1} a}{x_4} \end{aligned}

We recognise these expressions as multiples of the derivative of tan ⁡ − 1 ( X ) \tan^{-1}(X) evaluated at X = a X=a , which we know to be 1 1 + a 2 \frac{1}{1+a^2} . So L = 1 1 + a 2 [ 4 − 6 + 4 − 1 ] = 1 1 + a 2 L=\frac{1}{1+a^2} [4-6+4-1]=\boxed{\frac{1}{1+a^2}}

5 Helpful 4 Interesting 5 Brilliant 0 Confused

Thanks for the well explained solution. The limit indeed evaluates the derivative of tan ⁡ − 1 x \tan^{-1}{x} at x = a x=a . This is another way of computing derivatives using first principles. I derived this formula using Taylor series up to 4th order accuracy. The order of accuracy becomes relevant in the context of numerical differentiation.

Karan Chatrath - 3 months, 1 week ago

The general 4th order accurate formula (forward finite differences) for computing the derivative of a function is:

d f d x = lim ⁡ h → 0 − 25 f ( x ) + 48 f ( x + h ) − 36 f ( x + 2 h ) + 16 f ( x + 3 h ) − 3 f ( x + 4 h ) 12 h \frac{df}{dx}=\lim_{h \to 0} \frac{-25 f(x) + 48 f(x+h)- 36 f(x+2h) + 16 f(x+3h) - 3 f(x+4h)}{12h}

You can test this for other continuously differentiable functions to verify.

Karan Chatrath - 3 months, 1 week ago

Log in to reply

Aahh, I might have recognised that back when I was studying numerical analysis.

Chris Lewis - 3 months, 1 week ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...