How do you find the limit of #\lim _ { x \rightarrow 1} \frac { \sqrt { 5- x } - 2} { 1- x }#?

2 Answers
Noah G
Jun 17, 2017

The limit equals #1/4#.

Explanation:

If we start by evaluating directly, we get

#L = (sqrt(5 - 1) - 2)/(1 - 1) = (2 - 2)/(1 - 1) = 0/0#

Since we're in indeterminate form, we can use l'Hospitals rule, which states that #lim_(x-> a) f(x)/g(x) = lim_(x->a) (f'(x))/(g'(x))# only if #f(a)/g(a) = 0/0#.

The derivative of #sqrt(5 - x)# can be obtained using the chain rule.

#L = lim_(x->1) (-1/(2sqrt(5 - x)))/(-1)#

#L = lim_(x-> 1) 1/(2sqrt(5 - x))#

#L = 1/(2sqrt(5 - 1))#

#L = 1/4#

A graphical verification yields the same result. You should be able to see that at #x = 1#, the graph is at #y = 1/4#.

graph{y = (sqrt(5 - x) - 2)/(1 - x) [-10, 10, -5, 5]}

Hopefully this helps!

Ratnaker Mehta
Jun 17, 2017

# 1/4.#

Explanation:

The Reqd. Limit=#lim_(x to 1) (sqrt(5-x)-2)/(1-x),#

#=lim_(x to 1) (sqrt(5-x)-2)/(1-x) xx (sqrt(5-x)+2)/(sqrt(5-x)+2),#

#=lim_(x to 1) (5-x-4)/{(1-x)(sqrt(5-x)+2),#

#=lim_(x to 1) (cancel(1-x))/{(cancel(1-x))(sqrt(5-x)+2)},#

#=lim_(x to 1) 1/(sqrt(5-x)+2),#

#=1/(sqrt(5-1)+2),#

#=1/(2+2),#

#=1/4.#

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