Prove the following

Question:

$\lim _{x \rightarrow 1} \frac{x^{7}-2 x^{5}+1}{x^{3}-3 x^{2}+2}$

Solution:

This question can be easily solved using LH rule that is L. Hospital's rule which is given below

$\Rightarrow$ if $\lim _{x \rightarrow a g(x)} \frac{f(x)}{g}=\frac{0}{0}$ then $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f r(x)}{g t(x)}$

Given $\lim _{x \rightarrow 1} \frac{x^{7}-2 x^{5}+1}{x^{3}-3 x^{2}+2}$

If we apply the limit we will get in determinant form

$\lim _{x \rightarrow 1} \frac{x^{7}-2 x^{5}+1}{x^{3}-3 x^{2}+2}=\frac{0}{0}$

So now we have to apply L. Hospital's rule

$\Rightarrow$$\lim _{x \rightarrow 1} \frac{x^{7}-2 x^{5}+1}{x^{3}-3 x^{2}+2}=\lim _{x \rightarrow 1} \frac{\frac{d}{d x}\left(x^{7}-2 x^{5}+1\right)}{\frac{d}{d x}\left(x^{3}-3 x^{2}+2\right)}$

Now by differentiating we get

$\Rightarrow$$\lim _{x \rightarrow 1} \frac{7 x^{6}-10 x^{4}}{3 x^{2}-6 x}=\frac{7-10}{3-6}=\frac{-3}{-3}=1$

$\Rightarrow$$\lim _{x \rightarrow 1} \frac{x^{7}-2 x^{5}+1}{x^{2}-3 x^{2}+2}=1$

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