What is the limits?

Question

#lim(x->2)# #(sqrt(x+2)+x^2-6)#/#(x-2)#

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Answer 1 · 2018-03-27T18:44:28

The limit equals #17/4#

Use l'Hospitals rule, because we're in the #0/0# form.

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

Now we can readily evaluate.

#L = 1/(2sqrt(4)) + 4#

#L = 1/4 + 4 = 17/4#

The graph of the function confirms.

enter image source here

Hopefully this helps!

Answer 2 · 2018-03-27T19:11:24

#17/4#

Here,
#L=lim_(x to2)(sqrt(x+2)+x^2-6)/(x-2)#

#L=lim_(x to2)((x^2-6)+sqrt(x+2))/(x-2)xx((x^2-6)- sqrt(x+2))/((x^2-6)-sqrt(x+2))#

#=lim_(x to2)((x^2-6)^2-(sqrt(x+2))^2)/((x-2)((x^2-6)-sqrt(x+2)))#

#=lim_(x to2)1/((x^2-6)-sqrt(x+2))xxlim_(x to2)(x^4-12x^2+36-x- 2)/(x-2)#

#=1/(4-6-2)xxlim_(x to2)(x^4-4x^2-8x^2+16x-17x+34)/(x-2)#

#=-1/4lim_(x to2)[x^2(x^2-4)-8x(x-2)-17(x-2)]/(x-2)#

#=-1/4lim_(x to2)[x^2(x+2)-8x-17]#

#=-1/4(4(4)-16-17)#

#=17/4#