Question

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Ans these que fast plz


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Answer 1

Heya!!

lim { ( x + 1)^5 - 1 }/x

x–>0

put x = 0 we get 0/0. Now, use L'HOSPITALS RULE .

lim 5 ( x + 1 )⁴

x–>0

= 5

Q.2 :-

lim ( 1 - z³ ) × z³ / ( 1 - z^6 )

z–>1

put z = 1 we get 0/0. Now Use L'Hospitals rule .

lim ( 3z² - 6z^5 ) / ( -6z^5 )

z–>1

= ( 3 - 6 ) / -6

= 1/2

Answer 2

Answer:

$\lim_{x \to 0} \dfrac{(x+1)^5-1}{x}\\\\\text{Put x+1=y in the equation}\\\\\implies \lim_{y \to 1}\dfrac{y^5-1}{y-1}\\\\\bf{Using\:lim_{x\to a }\dfrac{x^n-a^n}{x-a}=na^{n-1}}\\\\\implies\mathsf{\lim_{y \to 1} 5\times 1^{5-1}}\\\\\implies 5\times 1^4\\\\\implies 5$

The answer is 5 .

$lim_{z\to 1}\dfrac{\dfrac{1}{z^3}-1}{\dfrac{1}{z^6}-1}\\\\\implies lim_{z\to 1}\dfrac{z^3(1-z^3)}{(1-z^6)}\\\\\implies lim_{z\to 1}\dfrac{z^3(1-z^3)}{(1+z^3)(1-z^3)}\\\\\implies lim_{z\to 1}\dfrac{z^3}{1+z^3}\\\\\implies \dfrac{1}{1+1}\\\\\implies \dfrac{1}{2}$

The answer is 1/2 .

Step-by-step explanation:

I was afraid that we had to use L'Hospital's rule when we see that both numerator and denominator are in indeterminate form .

But these limits can be solved by simplifying a little .

Use the power rules and rules of indices .

Only one expansion rule used is ( a + b )( a - b ) = a² - b²