Question

Please Answer these Question of Limits of Class 11​

Answer 1

ANSWER:

To Evaluate:

$\displaystyle\:\:\:\:\bullet\:\:\:\:\lim_{x\to1}\dfrac{1-x^{\frac{-1}{3}}}{1-x^{\frac{-2}{3}}}$

$\displaystyle\:\:\:\:\bullet\:\:\:\:\lim_{x\to-1}\dfrac{x^3+1}{x+1}$

Solution:

In the first part, we are given that,

$\displaystyle\implies\lim_{x\to1}\dfrac{1-x^{\frac{-1}{3}}}{1-x^{\frac{-2}{3}}}$

We can rewrite it as,

$\displaystyle\implies\lim_{x\to1}\dfrac{1-\dfrac{1}{x^{\frac{1}{3}}}}{1-\dfrac{1}{x^{\frac{2}{3}}}}$

Taking LCM in numerator and denominator,

$\displaystyle\implies\lim_{x\to1}\dfrac{\dfrac{x^{\frac{1}{3}}-1}{x^{\frac{1}{3}}}}{\dfrac{x^{\frac{2}{3}}-1}{x^{\frac{2}{3}}}}$

So, on simplifying,

$\displaystyle\implies\lim_{x\to1}\dfrac{x^{\frac{2}{3}}\left(x^{\frac{1}{3}}-1\right)}{x^{\frac{1}{3}}\left(x^{\frac{2}{3}}-1\right)}$

$\displaystyle\implies\lim_{x\to1}\dfrac{x^{\frac{2-1}{3}}\left(x^{\frac{1}{3}}-1\right)}{x^{\frac{2}{3}}-1}$

$\displaystyle\implies\lim_{x\to1}\dfrac{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)}{x^{\frac{2}{3}}-1}$

Now, we can rewrite x^(2/3) as, [x^(1/3)]^2.

That is,

$\displaystyle\implies\lim_{x\to1}\dfrac{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)}{\left(x^{\frac{1}{3}}\right)^2-1^2}$

Using, a^2 -b^2 = (a + b)(a - b),

$\displaystyle\implies\lim_{x\to1}\dfrac{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)}{\left(x^{\frac{1}{3}}+1\right) \left(x^{\frac{1}{3}}-1\right) }$

Now, x^(1/3) - 1, will be cancelled in the denominator as well as numerator. So,

$\displaystyle\implies\lim_{x\to1}\dfrac{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)}{\left(x^{\frac{1}{3}}+1\right) \left(x^{\frac{1}{3}}-1\right) }$

$\displaystyle\implies\lim_{x\to1}\dfrac{x^{\frac{1}{3}}}{x^{\frac{1}{3}}+1}$

Now, substituting the value of x = 1,

$\displaystyle\implies\dfrac{(1)^{\frac{1}{3}}}{(1)^{\frac{1}{3}}+1}$

We know that, value of 1^(1/3) = 1. So,

$\displaystyle\implies\dfrac{(1)^{\frac{1}{3}}}{(1)^{\frac{1}{3}}+1}$

$\displaystyle\implies\dfrac{1}{1+1}$

Hence,

$\displaystyle\implies\dfrac{1}{2}$

Therefore,

$\displaystyle\bf\implies\lim_{x\to1}\dfrac{1-x^{\frac{-1}{3}}}{1-x^{\frac{-2}{3}}}=\dfrac{1}{2}$

In the second part, we are given that,

$\displaystyle\implies\lim_{x\to-1}\dfrac{x^3+1}{x+1}$

We know that,

$\hookrightarrow (a+b)^3=(a+b)(a^2-ab+b^2)$

So,

$\displaystyle\implies\lim_{x\to-1}\dfrac{x^3+1}{x+1}$

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

Cancelling (x + 1),

$\displaystyle\implies\lim_{x\to-1} x^2-x+1$

Now, substituting the value of x = -1,

$\displaystyle\implies (-1)^2-(-1)+1$

$\displaystyle\implies 1+1+1$

Hence,

$\displaystyle\implies 3$

Therefore,

$\displaystyle\bf\implies\lim_{x\to-1}\dfrac{x^3+1}{x+1}=3$