Question

answer plz
lim u tends to 1 [u⁴-1÷u³-1]​

Answer 1

Solution:

Given Limit is:

$\displaystyle\rm = \lim_{u \to1} \bigg[ \dfrac{ {u}^{4} - 1 }{ {u}^{3} - 1 } \bigg]$

If we substitute u = 1, we get:

$\displaystyle\rm =\dfrac{ {1}^{4} - 1 }{ {1}^{3} - 1 }$

$\displaystyle\rm =\dfrac{0}{0}$

Which is an indeterminate form.

Consider the limit again:

$\displaystyle\rm = \lim_{u \to1} \bigg[ \dfrac{ {u}^{4} - 1 }{ {u}^{3} - 1 } \bigg]$

Can be written as:

$\displaystyle\rm = \lim_{u \to1} \bigg[ \dfrac{( {u}^{2} + 1)(u + 1)(u - 1)}{(u - 1)( {u}^{2} + u + 1)} \bigg]$

$\displaystyle\rm = \lim_{u \to1} \bigg[ \dfrac{( {u}^{2} + 1)(u + 1)}{ {u}^{2} + u + 1} \bigg]$

Now substitute u = 1 to get the value of the limit:

$\displaystyle\rm = \dfrac{( {1}^{2} + 1)(1 + 1)}{ {1}^{2} + 1+ 1}$

$\displaystyle\rm = \dfrac{2 \times 2}{3}$

$\displaystyle\rm = \dfrac{4}{3}$

$\displaystyle\rm = 1\dfrac{1}{3}$

Therefore:

$\displaystyle\rm \longrightarrow \lim_{u \to1} \bigg[ \dfrac{ {u}^{4} - 1 }{ {u}^{3} - 1 } \bigg] =1 \dfrac{1}{3}$

★ Which is our required answer.

Learn More:

Standard limits.

$\displaystyle\rm 1.\:\: \lim_{x\to0}\sin(x)=0$

$\displaystyle\rm 2.\:\: \lim_{x\to0}\cos(x)=1$

$\displaystyle\rm 3.\:\: \lim_{x\to0}\dfrac{\sin(x)}{x}=1$

$\displaystyle\rm 4.\:\: \lim_{x\to0}\dfrac{\tan(x)}{x}=1$

$\displaystyle\rm 5.\:\: \lim_{x\to0}\dfrac{1-\cos(x)}{x}=0$

$\displaystyle\rm 6.\:\: \lim_{x\to0}\dfrac{\sin^{-1}(x)}{x}=1$

$\displaystyle\rm 7.\:\: \lim_{x\to0}\dfrac{\tan^{-1}(x)}{x}=1$

$\displaystyle\rm 8.\:\: \lim_{x\to0}\dfrac{\log(1+x)}{x}=1$

$\displaystyle\rm 9.\:\: \lim_{x\to0}\dfrac{e^{x}-1}{x}=1$