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Answer:
![\lim_(x \to 5) = 7/2](https://img.qammunity.org/2021/formulas/mathematics/college/xuc9vuzjuvv4radw8sfakeq952ya2rxkfc.png)
Explanation:
We are given the equation:
![(x^(2)-3x-10 )/(2x-10)](https://img.qammunity.org/2021/formulas/mathematics/college/is1gzzqghtsu2iiyrlc3k0tbg16y2jxv2f.png)
Factor the numerator and denominator:
![((x - 5)(x+2) )/(2(x-5))](https://img.qammunity.org/2021/formulas/mathematics/college/zmt0ptebld9zvdpaupci2z7p64i6jibwc7.png)
'x - 5' is on both the numerator and denominator, so it gets cancelled out and becomes a "hole".
This means that at x = 5, there is a hole. There is a limit at x ⇒ 5. Find the hole by plugging 5 into the simplified equation:
= 7/2
Therefore:
![\lim_(x \to 5) = 7/2](https://img.qammunity.org/2021/formulas/mathematics/college/xuc9vuzjuvv4radw8sfakeq952ya2rxkfc.png)