Question

lim (x->0) x+ln (root over x2+1-x) /x3. Evalute​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{x + ln(\sqrt{x^{2} + 1} - x)}{x^{3} }$

As we know that,

Firstly, we put the value of x = 0 in the equation and check their indeterminant form, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{(0) + ln(\sqrt{(0)^{2}+ 1 } - 0)}{(0)^{3} }$

$\sf \implies \displaystyle \lim_{x \to 0} = \dfrac{0}{0} \ form$

As we can see that,

This equation is 0/0 indeterminant form.

Apply L-HOSPITAL'S rule in this equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{1 - \bigg( \dfrac{1}{\sqrt{x^{2} + 1} } \bigg)}{3x^{2} }$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{\sqrt{x^{2} + 1} - 1}{3x^{2} (\sqrt{x^{2} + 1} )}$

As we know that,

If roots exists in 0/0 form then we can simply rationalizes the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \bigg( \dfrac{\sqrt{x^{2} + 1} - 1}{3x^{2} (\sqrt{x^{2}+ 1 } )} \times \dfrac{\sqrt{x^{2} +1} + 1}{\sqrt{x^{2} + 1}+ 1 } \bigg)$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg( \dfrac{x^{2} + 1 - 1}{3x^{2} (\sqrt{x^{2} + 1} )(\sqrt{x^{2} + 1} + 1} \bigg)$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg( \dfrac{1}{3(\sqrt{x^{2} + 1} )(\sqrt{x^{2} + 1} + 1)} \bigg)$

Put the value of x = 0 in the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \bigg( \dfrac{1}{3 (\sqrt{(0)^{2} + 1} )(\sqrt{(0)^{2} + 1}+ 1 } \bigg)$

$\sf \implies \displaystyle \bigg( \dfrac{1}{3(1)(2)} \bigg) = \dfrac{1}{6}$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{x + ln(\sqrt{x^{2} + 1} - x)}{x^{3} } = \dfrac{1}{6}$

                                                                                                                   

MORE INFORMATION.

$\sf (1)= \displaystyle \lim_{x \to 0} \bigg( \dfrac{sin(x)}{x} \bigg) = \lim_{x \to 0} \bigg( \dfrac{x}{sin(x)} \bigg) = 1 ; \ \ \lim_{x \to 0} sin(x) = 0$

$\sf (2)= \displaystyle \lim_{x \to 0} cos(x) = \lim_{x \to0} \bigg( \dfrac{1}{cos(x)} \bigg) = 1.$

$\sf (3)= \displaystyle \lim_{x \to 0} \bigg( \dfrac{tan(x)}{x} \bigg) = \lim_{x \to 0} \bigg( \dfrac{x}{tan(x)} \bigg) = 1 \ ; \lim_{x \to 0} tan(x) = 0.$

$\sf (4)= \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin^{-1} x}{x} \bigg) = \lim_{x \to 0} \bigg( \dfrac{x}{sin^{-1} x} \bigg) = 1.$

$\sf (5)= \displaystyle \lim_{x \to 0} \bigg( \dfrac{tan^{-1} x}{x} \bigg) = \lim_{x \to 0} \bigg( \dfrac{x}{tan^{-1} x} \bigg) = 1.$

Answer 2

Answer:

□Given:-

• $\frac{lim}{x - 0} = \frac{x + in( \sqrt{ {x}^{2} + 1 } - x}{ {x}^{3} }$

□To find :-

• We should evaluate the equation and get answer.

♤Explanation :-

• Refer the attachment for more information.
• I'm this you will get correct answer. I cant do in in this answer so i am sharing the attachment .
• Here we get answer as 1/6.

♤Used formula :-

• Here we used L-Hospital Rule to get answer.

♧Hope it helps u mate .

♧Thank you .