giúp em bài giới hạn với ạ :(( lim x->3 (√ₓ₊₁₃ + √ᵪ₊₆ -7) ⁄ (ₓ²₋₉) ?

Đáp án:

\[\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 13}  + \sqrt {x + 6}  – 7}}{{{x^2} – 9}} = \frac{7}{{144}}\]

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 13}  + \sqrt {x + 6}  – 7}}{{{x^2} – 9}}\\
 = \mathop {\lim }\limits_{x \to 3} \left[ {\frac{{\sqrt {x + 13}  – 4}}{{{x^2} – 9}} + \frac{{\sqrt {x + 6}  – 3}}{{{x^2} – 9}}} \right]\\
 = \mathop {\lim }\limits_{x \to 3} \left[ {\frac{{\left( {\sqrt {x + 13}  – 4} \right)\left( {\sqrt {x + 13}  + 4} \right)}}{{\left( {{x^2} – 9} \right)\left( {\sqrt {x + 13}  + 4} \right)}} + \frac{{\left( {\sqrt {x + 6}  – 3} \right)\left( {\sqrt {x + 6}  + 3} \right)}}{{\left( {{x^2} – 9} \right)\left( {\sqrt {x + 6}  + 3} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 3} \left[ {\frac{{\left( {x + 13} \right) – {4^2}}}{{\left( {x – 3} \right)\left( {x + 3} \right)\left( {\sqrt {x + 13}  + 4} \right)}} + \frac{{\left( {x + 6} \right) – {3^2}}}{{\left( {x – 3} \right)\left( {x + 3} \right)\left( {\sqrt {x + 6}  + 3} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 3} \left[ {\frac{{x – 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)\left( {\sqrt {x + 13}  + 4} \right)}} + \frac{{x – 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)\left( {\sqrt {x + 6}  + 3} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 3} \left[ {\frac{1}{{\left( {x + 3} \right)\left( {\sqrt {x + 13}  + 4} \right)}} + \frac{1}{{\left( {x + 3} \right)\left( {\sqrt {x + 6}  + 3} \right)}}} \right]\\
 = \frac{1}{{\left( {3 + 3} \right)\left( {\sqrt {3 + 13}  + 4} \right)}} + \frac{1}{{\left( {3 + 3} \right).\left( {\sqrt {3 + 6}  + 3} \right)}}\\
 = \frac{7}{{144}}
\end{array}\)