It is given that `lim_(x->a)g(x) =oo` and `f(x)>= g(x)` . It has to be shown that `lim_(x->a) f(x)` is also equal to `oo` .

If `f(x)>= g(x)` , `f(x) - g(x) >= 0`

Let `lim_(x->a) f(x) = A` and `lim_x(->a) g(x) = B`

Let `epsi = A - B` . `lim_(x->a)(f(x) - g(x)) = A - B` . By the definition of limits for every `epsi > 0` , there exists a `delta > 0` such that for all values of x, `0 < |x - a| < delta => |f(x) - g(x) - A + B| < epsi` .

As `epsi = A - B` ,

`|f(x) - g(x) - A + B| < epsi`

=> `(f(x) - g(x) - A + B) < epsi`

=> `(f(x) - g(x) - A + B) < A - B`

=> `f(x) - g(x) < 2*(A - B)`

But as `f(x) - g(x)` is positive `2*(A - B)` is also positive or `A - B > 0`

=> `A > B`

=> `lim_(x->a) f(x) > lim_x(->a) g(x)`

As `lim_(x->a) g(x) = oo` this is possible only if `lim_(x->a) f(x) = oo`

**This proves that if `lim_(x->a)g(x) =oo` and `f(x)>= g(x)` , `lim_(x->a) f(x) = oo` **

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