No 5 ada yg bisa bantukah? Soal limit juga

Mapel : Matematika Wajib (Kelas XI SMA)
Materi : Kalkulus
Indikator : Mencari penyelesaian dan limit fungsi aljabar

Langkah Penyelesaian dan Jawaban :
[tex] \lim_{x \to \2} \frac{3- \sqrt{x+7} }{x^2+x-6} \\ \\ = \frac{3- \sqrt{x+7} }{x^2+x-6} \ . \frac{3+ \sqrt{x+7} }{3+ \sqrt{x+7} } \\ \\ = \frac{3^2 - ( \sqrt{x + 7} )^2}{(x+3)(x-2)(3+ \sqrt{x+7} )} \\ \\ = \frac{9 - (x+7)}{(x+3)(x-2)(3+ \sqrt{x+7})} \\ \\ = \frac{9 - x - 7}{(x+3)(x-2)(3+ \sqrt{x+7} )} \\ \\ = \frac{- x+2}{(x+3)(x-2)(3+ \sqrt{x+7} )} \\ \\ = \frac{- (x - 2)}{(x+3)(x-2)(3+ \sqrt{x+7} )} \\ \\ = - \frac{1}{(x+3)(3+ \sqrt{x+7} )} \\ \\ Substitusi \ 2 \\ \\ [/tex]

[tex]= - \frac{1}{(2+3)(3+ \sqrt{2+7} )} \\ \\ = - \frac{1}{5(3+ \sqrt{9} )} \\ \\ = - \frac{1}{5(3+3)} \\ \\ = - \frac{1}{30} [/tex]