Question

Class 11

limit
math
ncert questions

only right answer

Answer 1

GIVEN :-

$\implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \dfrac{x ^{2} - 16}{ \sqrt{x ^{2 } + 9} - 5}$

TO FIND :-

$\implies \: \displaystyle \bf \: value \: of \: \: \lim_{ x \: \to \: 4} \: \dfrac{x ^{2} - 16}{ \sqrt{x ^{2 } + 9} - 5}$

SOLUTION :-

$\implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \dfrac{x ^{2} - 16}{ \sqrt{x ^{2 } + 9} - 5}$

By RATIONALIZING the denominator,

$\implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \dfrac{x ^{2} - 16}{ \sqrt{x ^{2 } + 9} - 5} \: \: \times \frac{ \big( \sqrt{x ^{2} + 9} + 5 \big)} { \big(\sqrt{x ^{2} + 9 } + 5 \big) } \\ \\ \implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \dfrac{ \big(x ^{2} - 16 \big) \big( \sqrt{x ^{2} + 9 } + 5 \big)}{ \big(x ^{2} + 9 - 25 \big) } \\ \\ \implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \big( \sqrt{x ^{2} + 9 } + 5 \big) \\ \\ \implies \: \displaystyle \bf \big( \sqrt{16 + 9} + 5 \big) \\ \\ \implies \: \displaystyle \bf 5 + 5 \\ \\ \implies \boxed{ \red{ \bf \: 10}}$

Hence the Answer is 10.

ADDITIONAL INFORMATION :-

$\: \: \boxed{\begin{minipage}{4.5cm} \rm \displaystyle \: \lim_{n \to 0}\ sin\ x=0\\ \\ \rm \lim_{n \to 0}\ cos\ x=1\\\\\rm \lim_{x\to 0}\dfrac{1-cos\ x}{x}=0\\\\\rm \lim_{x\to \infty}\bigg(1+\dfrac{1}{x}\bigg)^x=e\\ \\ \rm \lim_{x\to a}\dfrac{(x^n-a^n)}{x-a}=n(a)^{n-1} \end{minipage}}$

Answer 2

$GIVEN :-</p><p></p><p>\implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \dfrac{x ^{2} - 16}{ \sqrt{x ^{2 } + 9} - 5}⟹x→4limx2+9−5x2−16</p><p></p><p>TO FIND :-</p><p></p><p>\implies \: \displaystyle \bf \: value \: of \: \: \lim_{ x \: \to \: 4} \: \dfrac{x ^{2} - 16}{ \sqrt{x ^{2 } + 9} - 5}⟹valueofx→4limx2+9−5x2−16</p><p></p><p>SOLUTION :-</p><p></p><p>\implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \dfrac{x ^{2} - 16}{ \sqrt{x ^{2 } + 9} - 5}⟹x→4limx2+9−5x2−16</p><p>By RATIONALIZING the denominator,</p><p>\begin{gathered} \implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \dfrac{x ^{2} - 16}{ \sqrt{x ^{2 } + 9} - 5} \: \: \times \frac{ \big( \sqrt{x ^{2} + 9} + 5 \big)} { \big(\sqrt{x ^{2} + 9 } + 5 \big) } \\ \\ \implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \dfrac{ \big(x ^{2} - 16 \big) \big( \sqrt{x ^{2} + 9 } + 5 \big)}{ \big(x ^{2} + 9 - 25 \big) } \\ \\ \implies \: \displaystyle \bf \lim_{ x \: \to \: 4} \: \big( \sqrt{x ^{2} + 9 } + 5 \big) \\ \\ \implies \: \displaystyle \bf \big( \sqrt{16 + 9} + 5 \big) \\ \\ \implies \: \displaystyle \bf 5 + 5 \\ \\ \implies \boxed{ \red{ \bf \: 10}}\end{gathered}⟹x→4limx2+9−5x2−16×(x2+9+5)(x2+9+5)⟹x→4lim(x2+9−25)(x2−16)(x2+9+5)⟹x→4lim(x2+9+5)⟹(16+9+5)⟹5+5⟹10</p><p>Hence the Answer is 10.</p><p></p><p>ADDITIONAL INFORMATION :-</p><p></p><p>$