How do you find the limit of #\lim _ { x \rightarrow 3} 5\cdot \frac { \sqrt { x + 1} - 2} { x - 3}#?

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Answer 1 · 2017-10-24T16:47:13

Because the expression evaluated at #x = 3#, yields the indeterminate form, #0/0#, one should use L'Hôpital's rule

To use L'Hôpital's rule, we differentiate both the numerator and the denominator:

#lim_(x to 3) 5((d(sqrt(x+1)-2))/dx)/((d(x-3))/dx)#

Compute the derivatives:

#lim_(x to 3) 5 (1/(2(sqrt(x+1))))/1#

Simplify and evaluate at #x = 3#:

#lim_(x to 3) 5/(2sqrt(x+1)) = 5/(2sqrt(3+1)) = 5/4#

L'Hôpital's rule states that, as goes the above limit, so goes the original limit:

#lim_(x to 3) 5(sqrt(x+1)-2)/(x-3) = 5/4#