Answer:
Hence, the limit of the function is:
1.5
Step-by-step explanation:
We have to find a reasonable estimate of limit of the given expression:
[tex]\lim_{x \to -1} \dfrac{x^6-1}{x^4-1}[/tex]
Since on putting the limit x=-1 we observe that the numerator and denominator both are equal to zero i.e we get a 0/0 form.Hence, we will apply L'hospitals rule to find the limit of the function.
We will firstly differentiate the numerator and denominator to obtain the limit.
On differentiating numerator we get:
[tex]6x^5[/tex]
and on differentiating denominator we obtain:
[tex]4x^3[/tex]
Hence, now we have to find the limit:
[tex]\lim_{x \to -1} \dfrac{6x^5}{4x^3}\\\\=\dfrac{6\times (-1)^5}{4\times (-1)^3}\\\\=\dfrac{6\times (-1)}[4\times (-1)}\\\\=\dfrac{-6}{-4}\\\\=\dfrac{6}{4}\\\\=\dfrac{3}{2}\\\\=1.5[/tex]
Hence, the limit of the function is:
1.5
