$$\displaystyle\lim_{{{x}\to{1}}}{\left({\frac{{{x}+{5}}}{{{x}+{2}}}}\right)}^{{4}}$$

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Brittany Patton
The given limit is:
$$\displaystyle\lim_{{{x}\to{1}}}{\left({\frac{{{x}+{5}}}{{{x}+{2}}}}\right)}^{{4}}$$
When substituting the direct limit, this limit will not undefined.
Therefore, to evaluate this take the limit directly.
That is,
$$\displaystyle\lim_{{{x}\to{1}}}{\left({\frac{{{x}+{5}}}{{{x}+{2}}}}\right)}^{{4}}={\left({\frac{{{1}+{5}}}{{{1}+{2}}}}\right)}^{{4}}$$
$$\displaystyle\lim_{{{x}\to{1}}}{\left({\frac{{{x}+{5}}}{{{x}+{2}}}}\right)}^{{4}}={\left({\frac{{{6}}}{{{3}}}}\right)}^{{4}}$$
$$\displaystyle\lim_{{{x}\to{1}}}{\left({\frac{{{x}+{5}}}{{{x}+{2}}}}\right)}^{{4}}={\left({2}\right)}^{{4}}$$
$$\displaystyle\lim_{{{x}\to{1}}}{\left({\frac{{{x}+{5}}}{{{x}+{2}}}}\right)}^{{4}}={16}$$
Answer: $$\displaystyle\lim_{{{x}\to{1}}}{\left({\frac{{{x}+{5}}}{{{x}+{2}}}}\right)}^{{4}}={16}$$