Question

Limit 12 th ..Maths easy ...​

Answer 1

Solution :

Now, $\displaystyle \lim_{x\to 1}\frac{1-\sqrt{x}}{1-x}$

$\displaystyle =\lim_{x\to 1}\frac{1-\sqrt{x}}{1^{2}-(\sqrt{x})^{2}}$

$\displaystyle =\lim_{x\to 1}\frac{1-\sqrt{x}}{(1+\sqrt{x})(1-\sqrt{x})}$

$\displaystyle =\lim_{x\to 1}\frac{1}{1+\sqrt{x}}$

$=\frac{1}{1+\sqrt{1}}$

$=\frac{1}{1+1}$

$=\frac{1}{2}$

$\to \boxed{\displaystyle \lim_{x\to 1}\frac{1-\sqrt{x}}{1-x}=\frac{1}{2}}$

Rules :

$\displaystyle 1.\:\lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$

$\displaystyle 2.\:\lim_{x\to \infty}(1+\frac{1}{x})^{n}=e$

$\displaystyle 3.\:\lim_{x\to 0}(1+x)^{\frac{1}{n}}=e$

Answer 2

Step-by-step explanation:

${\mathfrak{\red{\huge{\underline {Hola\:mate}}}}}$

Now, \displaystyle \lim_{x\to 1}\frac{1-\sqrt{x}}{1-x}

x→1

lim

1−x

1−

x

\displaystyle =\lim_{x\to 1}\frac{1-\sqrt{x}}{1^{2}-(\sqrt{x})^{2}}=

x→1

lim

1

2

−(

x

)

2

1−

x

\displaystyle =\lim_{x\to 1}\frac{1-\sqrt{x}}{(1+\sqrt{x})(1-\sqrt{x})}=

x→1

lim

(1+

x

)(1−

x

)

1−

x

\displaystyle =\lim_{x\to 1}\frac{1}{1+\sqrt{x}}=

x→1

lim

1+

x

1

=\frac{1}{1+\sqrt{1}}=

1+

1

1

=\frac{1}{1+1}=

1+1

1

=\frac{1}{2}=

2

1

\to \boxed{\displaystyle \lim_{x\to 1}\frac{1-\sqrt{x}}{1-x}=\frac{1}{2}}→

x→1

lim

1−x

1−

x

=

2

1

Rules :

\displaystyle 1.\:\lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}1.

x→a

lim

x−a

x

n

−a

n

=na

n−1

\displaystyle 2.\:\lim_{x\to \infty}(1+\frac{1}{x})^{n}=e2.

x→∞

lim

(1+

x

1

)

n

=e

\displaystyle 3.\:\lim_{x\to 0}(1+x)^{\frac{1}{n}}=e3.

x→0

lim

(1+x)

n

1

=e