Limiting Derivatives

Calculus Level 2

Evaluate :

$\lim _{x\rightarrow 5}\left ( \frac{d}{dx}\left ( \frac{x^{3}+1}{x^{2}+x} \right )\right )$

The answer is 0.96.

1 solution

Nihar Mahajan
Jan 23, 2016

Note that $\dfrac{x^3+1}{x^2+x} = \dfrac{(x+1)(x^2-x+1)}{x(x+1)} = \dfrac{x^2-x+1}{x}$

Using Quotient rule ,

$\begin{aligned} \dfrac{d}{dx} \dfrac{x^2-x+1}{x} &= \dfrac{x\dfrac{d}{dx} x^2-x+1 - (x^2-x+1)\dfrac{d}{dx} x }{x^2} \\ &= \dfrac{x(2x-1) - x^2+x-1}{x^2} \\ &= \dfrac{2x^2-x-x^2+x-1}{x^2} \\ &= \dfrac{x^2-1}{x^2} \\ &=1-\dfrac{1}{x^2} \end{aligned}$

Thus , $\displaystyle\lim_{x\rightarrow 5} 1-\dfrac{1}{x^2} = 1-\dfrac{1}{25} = \dfrac{24}{25} = \boxed{0.96}$

Nice solution. A slightly shorter version would result from noting that $\dfrac{x^{2} - x + 1}{x} = x - 1 + \dfrac{1}{x}$ , from which we quickly find the derivative to be $1 - \dfrac{1}{x^{2}}$ .

Brian Charlesworth - 5 years, 4 months ago

Yes! I have learnt quotient rule recently and hence I directly applied it as soon as I saw division. This teaches me not to haste , there is a short way possible ;)

Nihar Mahajan - 5 years, 4 months ago

Exactly ! ;)

A Former Brilliant Member - 5 years, 4 months ago

Congrats on learning differentiation bro! :)

Harsh Shrivastava - 5 years, 4 months ago

Thanks bro! I have started integration now , wait for me , I would catch upto you ;)

Nihar Mahajan - 5 years, 4 months ago

Sure! (This minimum word limit compels me to add more words in my comments)

Harsh Shrivastava - 5 years, 4 months ago

Last step should have x tends towards 5 isnt it ? @Nihar Mahajan

A Former Brilliant Member - 5 years, 4 months ago

Sorry , I have got a weird habit of writing infinity. Edited.

Nihar Mahajan - 5 years, 4 months ago

Nicely explained solution . (+1) ;)

A Former Brilliant Member - 5 years, 4 months ago

Limiting Derivatives

The answer is 0.96.

1 solution

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