Question

Evaluate
Limit. √x^2 +9-5 / x+4
0—>1

Answer 1

Solution:

Given limit:

$\displaystyle \rm = \lim_{x \to1} \: \dfrac{ \sqrt{ {x}^{2} + 9 - 5} }{x + 4}$

$\displaystyle \rm = \lim_{x \to1} \: \dfrac{ \sqrt{ {x}^{2} + 4} }{x + 4}$

Put x = 1 here to get the value of the limit.

$\displaystyle \rm = \dfrac{ \sqrt{ {1}^{2} + 4} }{1+ 4}$

$\displaystyle \rm = \dfrac{ \sqrt{5} }{5}$

$\displaystyle \rm = \dfrac{1 }{ \sqrt{5} }$

Therefore:

$\displaystyle \rm \longrightarrow\lim_{x \to1} \: \dfrac{ \sqrt{ {x}^{2} + 4} }{x + 4} = \dfrac{1}{ \sqrt{5} }$

★ Which is our required answer.

Answer:

$\displaystyle \rm \hookrightarrow\lim_{x \to1} \: \dfrac{ \sqrt{ {x}^{2} + 4} }{x + 4} = \dfrac{1}{ \sqrt{5} }$

Learn More:

Standard limits.

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

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

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

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

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

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

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

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

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